DaniWeb IT Discussion Community - Assembly

Trying to analyse code

bees1 — Sun, 08 Nov 2009 01:23:00 GMT

This is an example program that I am studying so that i can make my own mixed mode assembly/C program.

I am having trouble understanding on the assembly code, why I have to push SI twice at different times and only pop it out once. Also why push and pop DI when it's not used. I am confused from when I have push SI the second time.

/* An example to illustrate C function -> assembly -> a C function.
This program calls the assembly language procedure in file MARKS_A.ASM.
The program outputs min, max, and rounded average of two marks. */

 #include 

 int main(void)

 {             int mark1,mark2, min, max, av;

         int find_avg(int, int);

                extern  void stats(int, int, int*, int*, int*);

                scanf("%d%d",&mark1,&mark2);

                stats(mark1,mark2, &min, &max, &av);

        printf("Min = %d, Max = %d, Av = %d\n", min, max, av);

                return 0;   }





  /*********************************************************

     * Returns the rounded average required by the assembly 

    * procedure STATS in file MARKS_A.ASM.*/



        int find_avg(int total, int number)     {

                return((int)((double)total/number + 0.5)); }

;----------- --------- --------- --------- --------- --------- -

; Assembly program example to show call to a C function.

; This procedure receives a marks array and class size

; and returns minimum, maximum, and rounded average marks.

;----------- --------- --------- --------- --------- --------- -

.MODEL SMALL

EXTRN _find_avg:PROC

.CODE

PUBLIC _stats

_stats PROC 

push BP

mov BP,SP

push SI

push DI





mov AX,[BP+4] ; AX := mark1

mov DX,[BP+6] ; DX := mark2

cmp AX,DX ; put min into AX and max into DX

jl next

xchg AX,DX

next:

: mov SI,AX ; SI gets sum

add SI,DX



mov BX,[BP+8] ; return minimum

mov [BX],AX

mov BX,[BP+10] ; return maximum

mov [BX],DX 



; now call find_avg C function to compute average



mov AX,2 ;push number of marks

push AX

push SI ;push sum of marks

call _find_avg ; returns average in AX

add SP,4 ; clear stack

mov BX,[BP+12] ; return average

mov [BX],AX 

pop DI

pop SI

pop BP

ret ; NOTE calling program clears stack

_stats ENDP

END

Need help in factorial code

surfer2009 — Sat, 07 Nov 2009 10:53:33 GMT

; this code gereates maximum factorial of 8 decimal
; since, A is a 16-bit register and can hold max value of ;65535 dec (255h).
;

MOV A,#8d ; value for the factorial (1-8max)
MOV B,A ; B=A
MOV C,#1d ; C=1
here: ; LOOP
SUB B,C ; B-1
MUL AB ; AxB
CMP B,C ;if B=1
JNE here ;jump back

END

this code not work .please help me in thi and this also not work for lager values.(i have to take factorial of numbers 1 to 12)

x86 help with procedures

RobBrown — Sat, 07 Nov 2009 09:39:18 GMT

I am having a hard time getting my program to execute correctly. it is supposed to take two numbers entered by the user store them in an array, and reverse the array so when added they display correctly. i am to get the numbers using a procedure two seperate times, and reverse the arrays three times (reverse each entered number, and the sum array). furthermore when the user has entered one more than the max number of keys a + or = sign is to be entered for them. eg. i have the computer program set to take 3 numbers for each array so when the user enters a fourth a + or = sign should automatically appear. I cannot get the program to do this, or go any farther. I was hoping someone could let me know where i am going wrong with my code.

here it is:



           BOZOS_CODE SEGMENT    ;The "logical" segment is named BOZOS_CODE

             ASSUME CS:BOZOS_CODE   ;The "assume" statement tells the assembler

                                    ; that BOZOS_CODE is a code segment (CS)

             ORG 100h            ;In case we decide to convert the program

                                 ; to a COM file (makes 256 "empty" locations)



           START: jmp begin      ;Since this is a code segment the first non-

                                 ; directive item must be an instruction. We

                                 ; jump around our data area



          

          num_of_bytes EQU 3

          first_num db num_of_bytes dup (0), '$'

          second_num db num_of_bytes dup (0), '$'

          the_sum db num_of_bytes + 1 dup (0), '$'

          message_1 db 'Enter the first multi-digit number (up to 12 digits in length) followed by a'

                    db '"+" symbol then enter a second multi-digit number (up to 12 digits in '

                    db        'length) followed by a "=" symbol  $'

                  

                          



          key_stroke db 'x$'

          operand db 'x$'







GET_NUM:         mov ah, 00h    ;Read keyboard function

                 int 16h        ;Call BIOS (return occurs when key is pressed)

                                 ;AL now contains the ASCII code of key pressed

                 

                 mov key_stroke, al



                 mov ah, 09h                ;Display string function

                 mov dx, offset key_stroke  ;Pass the starting address of string

                 int 21h                    ;Call DOS

                 

                 mov al, key_stroke



                 cmp al, cl   

                 jne continue

                 ret                ;return



continue:           and al, 0fh



                 mov [BX], al

                 

                 inc BX

                

                 cmp [bx], ch



                 jbe get_num

                 

                 mov operand, cl

                

                 mov ah, 09h                  ;Display string function

                 mov dx, offset operand       ;Pass the starting address of string



                 int 21h          ;Call DOS

                 ret                  ;return



REV_NUM:          mov cl, [bx]



                 mov ch, [di]



                 mov [di], cl



                 mov [bx], ch



                 dec BX

                 inc di

                 

                 cmp di, bx

                                   

                 jae rev_num

                 ret

        

           

begin:                   mov ax, cs     

                  mov ds, ax     



                  mov cl, 0      

                  

                  mov ah, 06h    ;Scroll-up window function

                  mov al, 0      ; Clear entire window

                  mov bh, 1ch    ; Display "attribute byte" for cleared area

                                 ;  Blue background/Red characters

                                 ;  Bit definitions: Blink|Background R,G,B|

                                 ;                   Intensity|Character R,G,B

                  mov ch, 0      ; Y = 0     (Corner coordinates of window

                  mov cl, 0      ; X = 0      to clear)

                  mov dh, 24     ; Y = 24        

                  mov dl, 85     ; X = 79        

                  int 10h        ;Call BIOS



                                                   

                 mov ah, 02h    ;Set cursor position function

                 mov bh, 0      ; Display page 0

                 mov dh, 0      ; Y position  25 rows    (Y: 0 -> 24)

                 mov dl, 0      ; X position  80 columns (X: 0 -> 79)

                 int 10h        ;Call BIOS



                  

                 mov ah, 09h              ;Display string function

                 mov dx, offset message_1 ;Pass the starting offset of the

                                           ; string to be displayed

                                           ; (the string to be displayed starts 

                                           ;  at the label message (in the data

                                           ;  area) and ends with a '$')

                                           ;Note: DS must point to the beginning

                                           ;      of the segment containing the

                                           ;      string to be displayed, e.g.,

                                           ;      mov ax, cs  (usually done at

                                           ;      mov ds, ax   the beginning of

                                           ;                   the program)

                                           ;Note: This display procedure updates

                                           ;      the cursor position as each

                                           ;      character is displayed

                 int 21h                  ;Call DOS

                 

                 mov bx, offset first_num

                 mov cl, '+'

                 mov ch, num_of_bytes

                 

                 call get_num

                

                 mov bx, offset second_num

                 mov cl, '='

                 mov ch, num_of_bytes       ;do i need to do this again?

                

                 call get_num

                

                 mov di, offset first_num

                 

                 call rev_num

                

                 mov di, offset second_num

                 

                 call rev_num



                 mov BX, 0   ;Index set to zero now

                 mov ah, 0   ;ah now used for carry, and is set to zero



adder:           mov al, first_num[BX]

                 add al, second_num[BX]

                 add al, ah

                 

                 cmp al, 9

                 jbe no_carry

                 mov ah, 1 

                 sub al, 10

                 jmp make_ascii



no_carry:          mov ah, 0

                



make_ascii:         or al, 30h

                   

                 mov the_sum[BX], al

                 inc BX



                 cmp BX, num_of_bytes 

                 je fix_carry                ; if array is full jump to fix carry character

                    jmp adder

                 

fix_carry:         or ah, 30h

                

                 mov the_sum[BX], ah

                           

                 mov bx, num_of_bytes - 1

                 mov di, offset the_sum

                 

                 call rev_num 

                

                ;to display sum

                  

display:          mov ah, 09h                ;Display string function

                 mov dx, offset the_sum     ;Pass the starting address of string

                 int 21h                    ;Call DOS



                 ;To exit to DOS



exit:             mov ah, 4ch    ;Exit to DOS function

                  mov al, 0      ;ERRORLEVEL = 0

                  int 21h        ;Call DOS







           BOZOS_CODE ENDS

           END START           ;You MUST have a carriage return after START

thank you

int 21h/ah=09h string output

brando|away — Fri, 06 Nov 2009 22:50:10 GMT

im using fasm (flat assembler 1.68), and im trying to output a string with int21h/ah=09h

jmp main

msg db "hello world$"

main:

mov dx, msg

mov ah, 09

int 21h

this code outputs "hello world", but it outputs some control characters before it ([SOME WEIRD CONTROL CHARACTERS]hello world)

so how can i fix this?

Making a File Compressor. Need Help

leftovas17 — Fri, 06 Nov 2009 07:03:58 GMT

Here is the deal. I am writing a program in assembly that compresses and decompresses text based files. The user should be able to input a text file, and declare the name of what that file should be name when turned into a .drl file (just a random extension for my prog).

Right now I am working on reading the file to be compressed. First off I am reading the file one byte at a time and looking for " "(the space character). I want to search the text files for " word" and replace each instance of that word preceded with a space with a "key." This key will be designed using Huffman's Algorithm.

Anyways I want to read the character and as long as its not null, append it to the end of a string. How do I do that?(might be simple, but I'm not seeing it)

Next, how do I read from the file one byte prior to where I finished. Meaning: if I read a space and stop the read(because its the start of a new word) how do I re-read that space to add it to the word next time around?

Those are all the questions I am at the point of asking right now, but as I have seen plenty of times before, you guys are pretty strict on not giving out help on homework without effort shown. So here is my incomplete code. Feel free to make any and all helpful criticism.

Thank You.

Include Irvine32.inc  ;includes Master Irvine's Library



.data

message1 BYTE "Please type the file name of the file to be compressed",0     ;initiates the variable message1 and assigns it "Please type the file name of the file to be compressed", 0

message2 BYTE "Please type the filename to save the compression to(must end in .drl):",0       ;initiates the variable message2 and assigns it "Please type the filename to save the compression to:", 0

message3 BYTE "Cannot compress an empty file!",0                             ;initiates the variable message3 and assigns it "Cannot compress and empty file", 0

message4 BYTE "Please select one of the options:",0dh, 0Ah,

                                "1. Compress a file",0dh,0ah,

                                "2. DeCompress a file",0dh,0ah,0               ;initiates the variable message4 and assigns it "PLease select one 1. Compress 2. DeCompress", 0

message5 BYTE "Please type the file name of the file to be DeCompressed(must be a .drl file):",0     ;initiates the variable message1 and assigns it "Please type the file name of the file to be Decompressed", 0

message6 BYTE "Please type the filename to save the DeCompressed File to:",0       ;initiates the variable message2 and assigns it "Please type the filename to save the DeCompression to:", 0

handle1 DWORD ?                                                                    ;Uninitialized variable for handle1

handle2 DWORD ?                                                                    ;Uninitialized variable for handle2

BUFFER_SIZE = 1                                                                        ;sets buffer size to 1

buffer BYTE BUFFER_SIZE DUP(0)                                                        ;Duplicates buffer to buffer size and sets to 0

bytesRead DWORD ?                                                                ;Unintialized variable for bytesread

bytesWritten DWORD ?                                                                ;Unintialized variable for byteswritten

sourceFile BYTE 20 DUP(0),0                                                        ;initiates the variable for the sourceFile

destinationFile BYTE 20 DUP(0),0                                                ;initiates the variable for the destinationFile



.code

main PROC                                        ;beginning of main proc

call ShowPrompt                                ;calls showprompt proc

call checkForEmptyFile                ;calls checkforemptyfile proc

call getFileHandles                        ;calls getfilehandles proc

call ReadFromFile1                        ;calls read from file1 proc

call closeFiles                                ;calls the closeFiles proc

exit                                                ;properly exits

main ENDP                                        ;end of main proc



;--------------------------------------------------------------------------------------------------------------;

showPrompt PROC                                ;beginning of showPrompt proc

L1:                                                        ;loop one

mov edx, OFFSET message4        ;moves the offset of message 4 into edx

call WriteString                        ;displays message4

call crlf                                        ;new line

call ReadChar                                ;reads user input of a char

cmp al, "1"                                        ;compares input to the char '1'

je Compress                                        ;if equal then jump to Compress

cmp al, "2"                                        ;compares user input to the char '2'

je DeCompress                                ;if equal then jump to DeCompress

jmp L1                                                ;start loop over again if neither '1' nor '2'



Compress:                                        ;beginning of Compress label

mov edx, OFFSET message1        ;moves the offset of message1 into edx

call WriteString                        ;displays message1

call crlf                                        ;new line

mov edx, OFFSET sourceFile        ;moves the offset of sourceFile into edx

mov ecx, SIZEOF sourceFile        ;moves the sizeOf sourceFile into ecx

call ReadString                                ;reads the user input of a string

call crlf                                        ;new line

Compressb:                                        ;destination

mov edx, OFFSET message2        ;moves the offset of message2 into edx

call WriteString                        ;displays message2

call crlf                                                        ;new line

mov edx, OFFSET destinationFile                ;moves the offset of destinationFile into edx

mov ecx, SIZEOF destinationFile                ;moves the sizeOf destinationFile into ecx

call ReadString                                                ;reads the user input of a string

call crlf                                                        ;new line



mov edi, OFFSET destinationFile

mov eax, ".drl"

mov ecx, SIZEOF destinationFile

cld

repne scasb

jnz Compressb

dec edi

ret                                                                        ;return



DeCompress:                                                        ;beginning of DeCompress Label

mov edx, OFFSET message5                        ;moves the offset of message5 into ex

call WriteString                                        ;displays message5

call crlf                                                        ;new line

mov edx, OFFSET sourceFile                        ;moves the offset of sourceFile into edx

mov ecx, SIZEOF sourceFile                        ;moves the sizeOf sourceFile into ecx

call ReadString                                                ;reads the user input of a string

call crlf                                                        ;new line



mov edi, OFFSET sourceFile

mov eax, ".drl"

mov ecx, SIZEOF sourceFile

cld

repne scasb

jnz DeCompress

dec edi



mov edx, OFFSET message6                        ;moves the offset of message6 into edx

call WriteString                                        ;displays message6

call crlf                                                        ;new line

mov edx, OFFSET destinationFile                ;moves the offset of destinationFile into edx

mov ecx, SIZEOF destinationFile                ;moves the sizeOf destinationFile into ecx

call ReadString                                                ;reads the user input of a string

call crlf                                                        ;new line

ret                                                                        ;return

showPrompt ENDP                                                ;end of ShowPrompt proc



;--------------------------------------------------------------------------------------------------------------;

checkForEmptyFile PROC

mov edx, OFFSET sourceFIle                                ; moves the offset of sourceFile into edx

call OpenInputFile                                                ; calls openinputfile proc

cmp eax, INVALID_HANDLE_VALUE                        ;check validity of file handle

je Quit2

mov handle3, eax                                                                        ;jumps to quit label if invalid

mov edx, OFFSET buffer                                                ;moves the offset of buffer into edx

mov ecx, BUFFER_SIZE

mov eax, handle3                                                ;moves the buffer_Size into ecx

call ReadFromFile                                                ;calls ReadFromFile

mov bytesRead, eax                                                ;move eax into bytesRead

cmp eax, 0                                                        ;compare eax to 0

je QUIT                                                                ;if eax=0 then jump to quit

mov eax, handle3                                        

call Closefile                                                ;calls closeFiles proc                

ret                                                                        ;return



QUIT:                                                                ;beginning of QUIT label

mov edx, OFFSET message9                        ;moves the offset of message9 into edx

call WriteString                                        ;displays message9                                                        

exit                                                                ;properly exits



QUIT2:

mov edx, OFFSET message6                        ;moves the offset of message6 into edx

call WriteString                                        ;calls writestring

exit                                                                ;exits

checkForEmptyFile ENDP                                ;end of checkForEmptyFile proc



;--------------------------------------------------------------------------------------------------------------;

getFileHandles PROC                                                ;beginning of getfilehandles proc

mov edx, OFFSET sourceFIle                                ; moves the offset of sourceFile into edx

call OpenInputFile                                                ; calls openinputfile proc

cmp eax, INVALID_HANDLE_VALUE                        ;check validity of file handle

je Quit                                                                        ;jumps to quit label if invalid

mov handle1, eax                                                ;moves eax into handle1

mov edx, OFFSET destinationFile                                        ;moves the offset of the destination file into edx

call CreateOutputFile                                                ;calls the CreatOutputFile proc

cmp eax, INVALID_HANDLE_VALUE                        ;check validity of file handle

je Quit                                                                        ;jumps to quit label if invalid

mov handle2, eax                                                ;moves eax into handle2

ret                                                                                ;return



QUIT:                                                                        ;beginning of quit label

mov edx, OFFSET message6                                ;moves the offset of message6 into edx

call WriteString                                                ;writes edx

call crlf                                                                ;new line

call crlf                                                                ;new line

call main                                                                ;calls main proc

getFileHandles ENDP                                                        ;end of getFileHandles Proc



;--------------------------------------------------------------------------------------------------------------;

readFromFile1 PROC                                                        ;begining of readFromfile1 proc

Read:                                                                        ;beginning of read label

mov eax, handle1                                                ;moves handle1 into eax

mov edx, OFFSET buffer                                        ; moves the offset of the buffer into edx

mov ecx, BUFFER_SIZE                                        ;moves the buffer size into ecx

First:

call ReadFromFile

cmp eax, 0

je Quit2

cmp eax, " "

je Next





jmp New

NEXT:

mov bytesread, eax                                                ;moves the data into bytesread

New:

call ReadFromFile                                                ;reads data from file

cmp eax, 0                                                        ;compares eax to 0

je Quit2                                                        ;if eax = 0 then jump to Quit2

cmp eax, " "

jne NEXT

push eax                                                        ;push eax onto stack



pop eax                                                                ;pops eax off stack

cmp eax, 5                                                        ;compares eax to 5

jb Quit                                                                ;jump to quit if carry

call clear                                                        ;new line

jmp Read                                                        ;jump back to Read



Quit:                                                                ;beginning of Quit label

call crlf                                                        ;new line

mov edx, OFFSET message8                                        ;moves the offset of message8 into edx

call WriteString                                                ;displays message8

ret                                                                ;return





Quit2:                                                                ;beginning of Quit2 label

ret                                                                ;return

readFromFile1 ENDP                                                ;end of readFromFile1 proc



;--------------------------------------------------------------------------------------------------------------;

closeFiles PROC                                                        ;beginning of closeFiles proc

mov eax, handle1                                                ;moves handle1 into eax

call CloseFile                                                        ;closes the file

mov eax, handle2                                                ;moves handle2 into eax

call CloseFile                                                        ;close the file

ret                                                                ;return

closeFiles ENDP                                                        ;end of the closeFiles proc

Noob looking for guidance

dylank — Fri, 06 Nov 2009 04:01:02 GMT

Hi, I was using C++ to try to write objects to the screen, and decided to try this in a lower level language. I have searched the web and found some sample code:

name "vga"



; this program draws a tiny rectangle in vga mode. 



org  100h



jmp code



; dimensions of the rectangle: 

; width: 10 pixels 

; height: 5 pixels 



w equ 10

h equ 5





; set video mode 13h - 320x200 



code:   mov ah, 0

        mov al, 13h

        int 10h





; draw upper line: 



    mov cx, 100+w  ; column 

    mov dx, 20     ; row 

    mov al, 15     ; white 

u1: mov ah, 0ch    ; put pixel 

    int 10h



    dec cx

    cmp cx, 100

    jae u1



; draw bottom line: 



    mov cx, 100+w  ; column 

    mov dx, 20+h   ; row 

    mov al, 15     ; white 

u2: mov ah, 0ch    ; put pixel 

    int 10h



    dec cx

    cmp cx, 100

    ja u2



; draw left line: 



    mov cx, 100    ; column 

    mov dx, 20+h   ; row 

    mov al, 15     ; white 

u3: mov ah, 0ch    ; put pixel 

    int 10h



    dec dx

    cmp dx, 20

    ja u3





; draw right line: 



    mov cx, 100+w  ; column 

    mov dx, 20+h   ; row 

    mov al, 15     ; white 

u4: mov ah, 0ch    ; put pixel 

    int 10h



    dec dx

    cmp dx, 20

    ja u4





; pause the screen for dos compatibility: 



;wait for keypress 

  mov ah,00

  int 16h



; return to text mode: 

  mov ah,00

  mov al,03 ;text mode 3 

  int 10h



ret

Again, I did not write this, I found it at http://www.emu8086.com/dr/asm2html/a...phics.asm.html

My question is how would I go about compiling said code?

Use string to call variable

fitztho — Fri, 06 Nov 2009 03:54:45 GMT

I am looking to use a string to call a variable in assembly

I have the following:

msg_0      db   "Random phrase",10,0

msg_1      db   "Another Random Phrase",10

...(more messages)

msg_11    db  "Final Random Message",10,0 

msgIndex db  "msg_%d",0

then have the program push msgIndex and a random number to print one of the messages.

Is there a way that I can do this?

C and Assembly Mixed-Mode Hex comparison program

HypnotiqBIG — Thu, 05 Nov 2009 18:02:20 GMT

Hi, I have an assignment where i have to write a C and assembly mixed-mode program.

The c program takes in two characters which stand for two hex characters.
The assembly procedure then passes each character in turn to a C function which determines if the character is a valid ASCII HEX digit (namely 0-F).
If both characters are valid the assembly language procedure then compares the two HEX digits and returns the larger to the main program.
The main C program then prints out the larger digit.

I am unsure how to do no. 2, namely, how to determine if it's a valid ASCII HEX digit or just garbage. Suggestions?

Detect Multicore processors

group256 — Thu, 05 Nov 2009 17:35:30 GMT

Hey everyone,

For my project, I need to write an assembly/C code to detect multiprocessors. I really have no idea how as I have barely done assembly coding and I have no good reference or even idea how to start. Again, what I am going to do is, I'm trying to find how many processors on dual core or quad core chips there are and detect them and later on use them. Please help me out here.

Thanks a lot.

An os whith a menu.

Alex_ — Wed, 04 Nov 2009 06:52:25 GMT

hello people. I am doing an "os" (small) that's mainly a menu. It just prints some strings and highlights some words when you pres 1,2 or 3. And by pressing enter while a option is selected, it has to do something.
The problem is that the program gets stuck in the beginning ... and i don't know where is the problem. Can't run a debug because it's a bootable bin file. Can some one please tell me where is the error?
I'm using linux+ nasm

[BITS 16]      

jmp 07C0h:start

        snd     db  'Sound'

        lsnd        equ        $-snd

        abt        db        'About'

        labt        equ        $-abt

        hlp        db        'Help'

        lhlp        equ        $-hlp



    start:



is1:

        call init

        ;print "sound"

        mov        ax,1300h

        mov        bh,0

        mov        bl,2

        mov        cx,lsnd

        mov        dx,0

        push cs

        pop es

        mov bp,snd

        int 10h

wait_:

        mov        ah,00h

        int        16h



        cmp         ah,13

        je        make_sound



        cmp        ah,49

        je        is1



        cmp        ah,50

        je        is2



        cmp        ah,51

        je        is3



        jmp        wait_

is2:        

        call init

        ;print "About"

        mov        ax,1300h

        mov        bh,0

        mov        bl,2

        mov        cx,labt

        mov        dl,5

        push cs

        pop es

        mov bp,abt

        int 10h

        jmp        wait_



is3:        jmp        wait_

        

make_sound:

        mov        ax, 0E07h

        xor        bx, bx

        int        10h

        jmp        wait_

        

init:

        ;print "sound"

        mov        ax,1300h

        mov        bh,0

        mov        bl,5

        mov        cx,lsnd

        mov        dx,0

        push cs

        pop es

        mov bp,snd

        int 10h

        ;print "About"

        mov        ax,1300h

        mov        bh,0

        mov        cx,labt

        mov        dl,5

        push cs

        pop es

        mov bp,abt

        int 10h

        ;print "Help"

        mov        ax,1300h

        mov        bh,0

        mov        cx,lhlp

        mov        dl,10

        push cs

        pop es

        mov bp,hlp

        int 10h

        ret





    times 510-($-$$) db 0

    dw 0AA55h

The attachment shows the result of the execution. To me, it looks like he executes init then ;print "sound" then gets stuck. Maybe the function that i use to wait for a char isn't the right one?

Attached Images

Screenshot-1.png (17.5 KB)

Setting string colors in NASM

Goalatio — Wed, 04 Nov 2009 00:42:07 GMT

Hey all, I've been trying to get my int 10h to set the color for a STRING, not just a single character.

Here's my code, i'm running windows XP and using NASM16 assembler.

So, what I want it to do is display something like "Hello world", instead of just.. "a", etc.

the macro::

%macro color 3

mov cx,%1         ;CX value = amount of chars to print

mov bl,%2        ;BL value = HEXI color attr

mov al,%3        ;AL value = character to print

mov ah,9        ;Function 9, display char

int 10h                ;Call the SCREEN vector



%endmacro

the code::

[org 0100h]

%include "Library.asm"



[section .text]



color 1,6ah,hello     ;Use library to attempt to display string

string eol

exit 0



[section .data]

hello db "Hello, world.", 13, 10, "$"

eol db 13, 10, "$"

Any help would be great, thanks.

Needing some guidance

RayvenHawk — Tue, 03 Nov 2009 18:06:28 GMT

To be upfront and all, this is indeed a homework assignment. But I'm not looking for anyone to write the code I need just help me with a few functions I'm needing to use (the book is VERY vague on how to use these instructions).

Assignment is open an encrypted file, decrypt it and then send it to a new text file.

Ok so far I have this

.data

BUFFER_SIZE = 5000

buffer BYTE BUFFER_SIZE DUP(?)

filename BYTE "myfile.txt",0 (this is coming from my book)



.code

mov edx,OFFSET filename

call OpenInputFile

mov edx,OFFSET buffer

mov ecx,BUFFER_SIZE

call ReadFromFile

call WriteToFile

As I said the book is extremly vague on how these instructions truly work. My question is what is the normal procedure to open a file, read it and then output it to a new file?

Define Type of Assembly

Fong~ — Tue, 03 Nov 2009 10:14:24 GMT

Can anyone tell me what the assembly type for bottom assembly program?

#DEFINE PAGE0 BCF $03,5
#DEFINE PAGE1 BSF $03,5

INDF: .EQU $00 ;page 0, 1, 2, 3
TMR0: .EQU $01 ;page 0, 2
OPTION: .EQU $01 ;page 1, 3
PCL: .EQU $02 ;page 0, 1, 2, 3
STATUS: .EQU $03 ;page 0, 1, 2, 3
FSR: .EQU $04 ;page 0, 1, 2, 3

PORTA: .EQU $05 ;page 0
TRISA: .EQU $05 ;page 1
PORTB: .EQU $06 ;page 0, 2
TRISB: .EQU $06 ;page 1, 3
PORTC: .EQU $07 ;page 0
TRISC: .EQU $07 ;page 1
PORTD: .EQU $08 ;page 0
TRISD: .EQU $08 ;page 1
PORTE: .EQU $09 ;page 0
TRISE: .EQU $09 ;page 1

INTCON: .EQU $0B ;page 0, 1, 2, 3
ADCON0: .EQU $1F ;page 0
ADCON1: .EQU $1F ;page 1

LOOPA: .EQU $20 ;loop used by LCD send routine
LOOPB: .EQU $21 ;general loop
STORE: .EQU $22 ;general store
STORE1: .EQU $23 ;general store
STORE2: .EQU $24 ;general store
STORE3: .EQU $25 ;general store
RSLINE: .EQU $26 ;LCD function flag store
SWITCH: .EQU $27 ;switch status store
MARK1: .EQU $28 ;freq count polarity marker 1
MARK2: .EQU $29 ;freq count polarity marker 2

DEC1: .EQU $2A ;decimalisation byte 1
DEC2: .EQU $2B ; byte 2
DEC3: .EQU $2C ; byte 3
DEC4: .EQU $2D ; byte 4
DEC5: .EQU $2E ; byte 5
DEC6: .EQU $2F ; byte 6 (dummy)

ANSA1: .EQU $30 ;decimalisation answer store 1
ANSA2: .EQU $31 ; answer 2
ANSA3: .EQU $32 ; answer 3
ANSA4: .EQU $33 ; answer 4
ANSA5: .EQU $34 ; answer 5

FREQ0: .EQU $35 ;frequency counter lsb
FREQ1: .EQU $36 ;frequency counter nsb
FREQ2: .EQU $37 ;frequency counter nsb
FREQ3: .EQU $38 ;frequency counter nsb
FREQ4: .EQU $39 ;frequency counter msb
CLKCNT: .EQU $3A ;timing counter
SLOWIT: .EQU $3B ;delay factor for PAUSE
LOOPX: .EQU $3C ;delay loop for LCDOUT

OUT7: .EQU $3D ;8 bytes for freq store and output to LCD
OUT6: .EQU $3E
OUT5: .EQU $3F
OUT4: .EQU $40
OUT3: .EQU $41
OUT2: .EQU $42
OUT1: .EQU $43
OUT0: .EQU $44
LCDSTOR1: .EQU $45
LCDSTOR2: .EQU $46
STOREC: .EQU $47
STORED: .EQU $48
TENFLG: .EQU $49
TENCNT: .EQU $4A
RANGE: .EQU $4B ;frequency cap selection
SHAPE: .EQU $4C ;waveform shape selection
SAMPLE: .EQU $4D ;1sec/10sec flag
CLKVAL1: .EQU $4E
CLKVAL2: .EQU $4F

;extends to $7F (max limit)

W: .EQU 0
F: .EQU 1
C: .EQU 0
DC: .EQU 1
Z: .EQU 2

RP0: .EQU 5 ;STATUS reg
RP1: .EQU 6 ;STATUS reg
GIE: .EQU 7 ;INTCON reg

.ORG $0004 ;Interrupt vector address
GOTO START ;Jump to interrupt routine on interrupt
.ORG $0005 ;Start of program memory

clrf INTCON

GIEOFF: BCF INTCON,GIE ;turn off global interrupts
BTFSC INTCON,GIE
goto GIEOFF

goto START

TABLCD: addwf PCL,F ;LCD initialisation table
retlw %00110011 ;initialise lcd - first byte
retlw %00110011 ;2nd byte (repeat of first)
retlw %00110010 ;set for 4-bit operation
retlw %00101100 ;set for 2 lines
retlw %00000110 ;set entry mode to increment each address
retlw %00001100 ;set display on, cursor off, blink off
retlw %00000001 ;clear display
retlw %00000010 ;return home, cursor & RAM to zero
;end initialisation table

TBDEC1: ;table for decimalisation lsb
addwf PCL,F ;add program counter
retlw $10 ;lsb of 10000
retlw $E8 ;lsb of 1000
retlw $64 ;lsb of 100
retlw $0A ;lsb of 10

TBDEC2: ;table for decimalisation msb
addwf PCL,F ;add program counter
retlw $27 ;msb of 10000
retlw $03 ;msb of 1000
retlw 0 ;msb of 100
retlw 0 ;msb of 10

RANGIT: addwf PCL,F ;range control vals
retlw %11111111
retlw %11111101
retlw %11111011
retlw %11110111
retlw %11101111
retlw %11011111
retlw %10111111
retlw %01111111

WAVEIT: addwf PCL,F ;waveform message routing
goto SQUARE
goto TRNGLE
goto SINE

SINE: movf LOOPA,W
addwf PCL,F ;sine message route
retlw 'S'
retlw 'I'
retlw 'N'
retlw 'E'

SQUARE: movf LOOPA,W
addwf PCL,F ;square message route
retlw 'S'
retlw 'Q'
retlw 'R'
retlw ' '

TRNGLE: movf LOOPA,W
addwf PCL,F ;triangle message route
retlw 'T'
retlw 'R'
retlw 'I'
retlw ' '

;..............

START: bcf STATUS,RP0
bcf STATUS,RP1
clrf PORTA
clrf PORTB
clrf PORTC
clrf PORTD
clrf PORTE
PAGE1
clrf TRISA ;PORTA as output
movlw %00111111
movwf TRISB ;PORTB as input
movlw 255
movwf TRISC ;PORTC as input
movlw %11001111
movwf TRISD ;RD0-RD3, RD6-RD7 as input, RD4-RD5 as output
movlw %00000111
movwf TRISE ;PORTE as input
movlw %00000111 ;set LHS justify, RA0-RA3 as digital
movwf ADCON1
movlw %10000110 ;PORTB pullups off, timer 1:128 (1/25th sec)
movwf OPTION
PAGE0

call PAUSIT
LCDSET: clrf LOOPB
clrf RSLINE
LCDST2: movf LOOPB,W
call TABLCD
call LCDOUT
incf LOOPB,F
btfss LOOPB,3
goto LCDST2
call PAUSIT

clrf MARK2
clrf SWITCH
clrf FREQ0
clrf FREQ1
clrf FREQ2
clrf FREQ3
clrf FREQ4
clrf ANSA1
clrf ANSA2
clrf ANSA3
clrf ANSA4
clrf ANSA5
clrf DEC1
clrf DEC2
clrf DEC3
clrf DEC4
clrf DEC5
clrf DEC6
clrf TENFLG
clrf TENCNT
clrf MARK1
clrf MARK2
clrf RANGE
clrf SAMPLE
movlw 0 ;......2
movwf SHAPE
call WAVE1
movlw 25
movwf CLKVAL1
movlw 250
movwf CLKVAL2

movf CLKVAL1,W
movwf CLKCNT

bsf PORTD,4 ;reset counter IC6
bcf PORTD,4
bsf PORTD,5 ;enable clock input

movlw 4
movwf RANGE
bcf INTCON,2

goto SHWRNG

;............................ END OF SETUP

MAIN: btfss INTCON,2
goto MF1
bcf INTCON,2

movf PORTE,W ;is a range switch pressed?
btfss STATUS,Z
goto RANGESW ;yes
btfsc PORTB,0 ;is waveshape switch pressed?
call WAVEFRM ;yes
decfsz CLKCNT,F ;dec timing counter, is it 0
goto MF1 ;no
goto GETFREQ ;yes

MF1: call INCCNT
goto MAIN

INCCNT: btfss PORTC,0
goto MF2
bsf MARK2,0
return

MF2: btfss MARK2,0 ;is MARK2 = 1?
return ;no
bcf MARK2,0
incfsz FREQ2,F ;yes, inc counters accordingly
return ;(responds to negative-going aspect of IC6 Q12)
incfsz FREQ3,F
return
incf FREQ4,F
return

;............show freq routine

GETFREQ: bcf PORTD,5 ;turn off clock input
nop
nop
call INCCNT ;*****
movf PORTD,W ;get pins which include Q0-Q3
movwf STORED
movf PORTC,W
movwf STOREC
bsf PORTD,5 ;turn on clock input

bcf STATUS,C ;shift freq counters right by 4 places
rrf FREQ4,F
rrf FREQ3,F
rrf FREQ2,F
rrf FREQ1,F
bcf STATUS,C
rrf FREQ4,F
rrf FREQ3,F
rrf FREQ2,F
rrf FREQ1,F
bcf STATUS,C
rrf FREQ4,F
rrf FREQ3,F
rrf FREQ2,F
rrf FREQ1,F
bcf STATUS,C
rrf FREQ4,F
rrf FREQ3,F
rrf FREQ2,F
rrf FREQ1,F

btfsc STORED,1 ;Q1
bsf FREQ0,0
btfsc STORED,2 ;Q2
bsf FREQ0,1
btfsc STORED,3 ;Q3
bsf FREQ0,2
btfsc STOREC,4 ;Q4
bsf FREQ0,3
btfsc STOREC,6 ;Q5
bsf FREQ0,4
btfsc STOREC,7 ;Q6
bsf FREQ0,5
btfsc STOREC,5 ;Q7
bsf FREQ0,6
btfsc STOREC,3 ;Q8
bsf FREQ0,7

btfsc STORED,0 ;Q9
bsf FREQ1,0
btfsc STOREC,2 ;Q10
bsf FREQ1,1
btfsc STOREC,1 ;Q11
bsf FREQ1,2
btfsc STOREC,0 ;Q12
bsf FREQ1,3

movlw FREQ0
call DECIML ;decimalise digits 1 to 4
movf DEC1,W ;copy answer into OUT-OUT3
movwf OUT0
movf ANSA1,W
movwf OUT1
movf ANSA2,W
movwf OUT2
movf ANSA3,W
movwf OUT3

GF3: movf ANSA4,W ;copy overflow val into prev freq positions
movwf FREQ0
movf ANSA5,W
movwf FREQ1
clrf FREQ2
clrf FREQ3
clrf FREQ4
movlw FREQ0
call DECIML ;now decimal digits 5 to 8

movf ANSA3,W ;copy answer into OUT4 to OUT7
movwf OUT7
movf ANSA2,W
movwf OUT6
movf ANSA1,W
movwf OUT5
movf DEC1,W
movwf OUT4

GF4: call LCD5 ;output freq data to LCD
bsf RSLINE,4
clrf MARK1 ;clear leading zero flag

movlw OUT7
movwf FSR
movlw 8
movwf LOOPA
MAINA: movf INDF,W
btfss STATUS,Z ;is val = 0?
goto MAINB ;no
btfsc MARK1,0 ;has val >0 been output?
goto MAINB ;yes
movf LOOPA,W ;is it the last byte of the loop?
xorlw 1
btfsc STATUS,Z
goto MAINB ;yes
movlw ' ' ;clear leading zero
goto MAINC

MAINB: bsf MARK1,0 ;format as ASCII decimal val
andlw 15
iorlw 48

MAINC: call LCDOUT
incf FSR,F
decfsz LOOPA,F
goto MAINA
btfss SAMPLE,0
goto GF5
call LCD12
bsf RSLINE,4
movlw '.'
call LCDOUT

decf FSR,F
movf INDF,W
andlw 15
iorlw 48
call LCDOUT
goto GF6

GF5: movlw ' '
call LCDOUT

GF6: movlw 'H'
call LCDOUT
movlw 'z'
call LCDOUT

MAINX: clrf FREQ0 ;clear freq counters
clrf FREQ1
clrf FREQ2
clrf FREQ3
clrf FREQ4
clrf MARK2

movf CLKVAL1,W
btfsc SAMPLE,0 ;is 10sec rate on?
movf CLKVAL2,W
movwf CLKCNT
bsf PORTD,4 ;reset counter IC6
bcf PORTD,4
bsf PORTD,5 ;enable clock input

clrf TMR0
bcf INTCON,2 ;clear interrupt

goto MAIN ;start sampling next batch

;.............

PAUSIT: movlw 7
movwf SLOWIT
bcf INTCON,2
PAUSE: btfss INTCON,2
goto PAUSE
bcf INTCON,2
decfsz SLOWIT,F
goto PAUSE
return

;................

LCD1: movlw %10000000 ;LCD cell position codes - not all used
goto LCDLIN
LCD5: movlw %10000101
goto LCDLIN
LCD6: movlw %10000110
goto LCDLIN
LCD7: movlw %10000111
goto LCDLIN

LCD9: movlw %10001001
goto LCDLIN
LCD12: movlw %10001100
goto LCDLIN

LCD21: movlw %11000000
goto LCDLIN
LCD24: movlw %11000100
goto LCDLIN
LCD25: movlw %11000101
goto LCDLIN
LCD26: movlw %11000110
goto LCDLIN
LCD27: movlw %11000111
goto LCDLIN

LCD28: movlw %11001000
goto LCDLIN
LCD29: movlw %11001001
goto LCDLIN

LCDLIN: BCF RSLINE,4

LCDOUT: movwf STORE
movlw 50
movwf LOOPX
DELAY: decfsz LOOPX,F
goto DELAY
call SENDIT
SENDIT: swapf STORE,F
movf STORE,W
andlw 15
iorwf RSLINE,W
movwf LCDSTOR1
clrf LCDSTOR2
movlw 6
movwf LOOPX

SHIFT: bcf STATUS,C ;rearrange bits to suit PORTA pin order
rrf LCDSTOR1,F
rlf LCDSTOR2,F
decfsz LOOPX,F
goto SHIFT
movf LCDSTOR2,W
movwf PORTA
bsf PORTA,0 ;toggle LCD line E
bcf PORTA,0
bsf PORTA,5 ;turn on IC2 sync output line
return

;..............convert binary to hex - not used in final model

LCDHEX: movwf STORE2 ;split & format decimal byte as HEX for LCD
swapf STORE2,W ;get tens nibble
andlw 15
movwf STORE1
addlw 6
btfss STATUS,DC
goto HEX2
movf STORE1,W
addlw 55 ;set as alpha
goto HEX3
HEX2: movf STORE1,W
iorlw 48 ;set as numeral
HEX3: call LCDOUT ;send it
movf STORE2,W ;get units
andlw 15
movwf STORE1
addlw 6
btfss STATUS,DC
goto HEX4
movf STORE1,W
addlw 55 ;set as alpha
goto HEX5
HEX4: movf STORE1,W
iorlw 48 ;set as numeral
HEX5: call LCDOUT
return

;........... used to convert binary bytes to decimal for LCD

DECIML: ;decimalise binary number
;copy source into working area at DEC
;source address is brought in on W
;DEC1-5 becomes source
;answer goes into ANSA
call COPYFD ;copy data at FSR into DEC
clrf STORE1
clrf STORE3
clrf ANSA5
movlw ANSA4 ;set answer store address
movwf FSR
clrf LOOPA

DCML0: movf LOOPA,W
call TBDEC1 ;subtract lsb from source value
subwf DEC1,F
btfsc STATUS,C ;is there a borrow?
goto DCML1 ;no, so sub msb
movlw 1
subwf DEC2,F ;yes so decrement next byte
btfsc STATUS,C ;is there a borrow?
goto DCML1 ;no, so sub msb
movlw 1 ;yes
subwf DEC3,F
btfsc STATUS,C ;is there a borrow?
goto DCML1 ;no, so sub msb
movlw 1 ;yes
subwf DEC4,F
btfsc STATUS,C ;is there a borrow?
goto DCML1 ;no, so sub msb
movlw 1 ;yes
subwf DEC5,F
btfsc STATUS,C ;is there a borrow?
goto DCML1 ;no, so sub msb

DCML3: movf LOOPA,W ;yes, so re-add last table values lsb
call TBDEC1
addwf DEC1,F
btfss STATUS,C ;is there a carry?
goto DCML5 ;no so exit loop
incf DEC2,F ;yes so inc next byte
btfss STATUS,Z ;is there a carry? (is it zero)
goto DCML5 ;no so exit loop
incf DEC3,F ;yes
btfss STATUS,Z ;is there a carry? (is it zero)
goto DCML5 ;no so exit loop
incf DEC4,F ;yes
btfss STATUS,Z ;is there a carry? (is it zero)
goto DCML5 ;no so exit loop
incf DEC5,F ;yes
goto DCML5 ;exit loop

DCML1: movf LOOPA,W
call TBDEC2 ;subtract msb from source value
subwf DEC2,F
btfsc STATUS,C ;is there a borrow?
goto DCML2 ;no
movlw 1
subwf DEC3,F ;yes so decrement next byte
btfsc STATUS,C ;is there a borrow?
goto DCML2 ;no
movlw 1
subwf DEC4,F ;yes so decrement next byte
btfsc STATUS,C ;is there a borrow?
goto DCML2 ;no
movlw 1
subwf DEC5,F ;yes so decrement next byte
btfss STATUS,C ;is there a borrow?
goto DCML4 ;yes

DCML2: incfsz STORE1,F ;inc counter & continue looping till zero
goto DCML0
incf STORE3,F
goto DCML0

DCML4: movf LOOPA,W ;re-add last table values lsb
call TBDEC1
addwf DEC1,F
btfss STATUS,C ;is there a carry?
goto DCML4A ;no so add msb
incf DEC2,F ;yes so inc next byte
btfss STATUS,Z ;is there a carry? (is it zero)
goto DCML4A ;no so add msb
incf DEC3,F ;yes, so inc next byte
btfss STATUS,Z ;is there a carry? (is it zero)
goto DCML4A ;no so add msb
incf DEC4,F ;yes, so inc next byte
btfss STATUS,Z ;is there a carry? (is it zero)
goto DCML4A ;no so add msb
incf DEC5,F ;yes, so inc next byte

DCML4A: movf LOOPA,W ;re-add last table values msb
call TBDEC2
addwf DEC2,F
btfss STATUS,C ;is there a carry?
goto DCML5 ;no
incf DEC3,F ;yes, so inc next byte
btfss STATUS,Z ;is there a carry? (is it zero)
goto DCML5 ;no
incf DEC4,F ;yes, so inc next byte
btfss STATUS,Z ;is there a carry? (is it zero)
goto DCML5 ;no
incf DEC5,F ;yes, so inc next byte

DCML5: movf STORE3,W
btfss STATUS,Z
movwf ANSA5
movf STORE1,W ;store counter and loop for next values
movwf INDF ;INDF here is ANSA group

clrf STORE1
clrf STORE3
decf FSR,F

incf LOOPA,F
btfss LOOPA,2 ;is it = 4?
goto DCML0 ;no

DCML6: return

COPYFD: movwf FSR ;copy data at FSR to DEC
movf INDF,W
movwf DEC1
incf FSR,F
movf INDF,W
movwf DEC2
incf FSR,F
movf INDF,W
movwf DEC3
incf FSR,F
movf INDF,W
movwf DEC4
incf FSR,F
movf INDF,W
movwf DEC5
clrf DEC6
return

;...........range switch

RANGESW: btfsc PORTE,2
goto SAMRATE
call OK

btfsc PORTE,1
goto DECRNG

INCRNG: incf RANGE,W
goto SHWRNG
DECRNG: decf RANGE,W

SHWRNG: andlw 7
movwf RANGE
call LCD21
bsf RSLINE,4
movlw 'R'
call LCDOUT
movlw 'a'
call LCDOUT
movlw 'n'
call LCDOUT
movlw 'g'
call LCDOUT
movlw 'e'
call LCDOUT

movf RANGE,W
sublw 7

iorlw 48
call LCDOUT
movf PORTB,W
andlw %11000000
movwf PORTB

bcf PORTD,6
bcf PORTD,7

movf RANGE,W
call RANGIT
andlw %00111111
PAGE1
movwf TRISB
PAGE0
movf RANGE,W
call RANGIT
andlw %11000000
iorlw %00001111
PAGE1
movwf TRISD
PAGE0
call PAUSIT
call PAUSIT
movf PORTE,W
andlw 3
btfss STATUS,Z
goto RANGESW
goto SAM1

;...........set sample rate 1sec/10sec

SAMRATE: incf SAMPLE,F
bcf SAMPLE,1
call OK

SAM1: clrf FREQ0 ;clear freq counters
clrf FREQ1
clrf FREQ2
clrf FREQ3
clrf FREQ4
clrf MARK2

call LCD27
bsf RSLINE,4
movlw 'T'
call LCDOUT
movlw 'i'
call LCDOUT
movlw 'm'
call LCDOUT
movlw 'e'
call LCDOUT
movlw '1'
call LCDOUT
movlw ' '
btfsc SAMPLE,0
movlw '0'
call LCDOUT
movlw 's'
call LCDOUT
movlw 'e'
call LCDOUT
movlw 'c'
call LCDOUT

SAM3: movf PORTE,W
andlw 7
btfss STATUS,Z
goto SAM3

call LCD5 ;output freq data to LCD
bsf RSLINE,4
movlw ' '
call LCDOUT
movlw ' '
call LCDOUT
movlw 'C'
call LCDOUT
movlw 'O'
call LCDOUT
movlw 'U'
call LCDOUT
movlw 'N'
call LCDOUT
movlw 'T'
call LCDOUT
movlw 'I'
call LCDOUT
movlw 'N'
call LCDOUT
movlw 'G'
call LCDOUT
movlw ' '
call LCDOUT

movf CLKVAL1,W
btfsc SAMPLE,0
movf CLKVAL2,W
movwf CLKCNT

bsf PORTD,4 ;reset counter IC6
bcf PORTD,4

goto MAIN

OK: call LCD5 ;output freq data to LCD
bsf RSLINE,4
movlw ' '
call LCDOUT
movlw ' '
call LCDOUT
movlw 'O'
call LCDOUT
movlw 'K'
call LCDOUT

movlw 7
movwf LOOPA
OK2: movlw ' '
call LCDOUT
decfsz LOOPA,F
goto OK2
movlw '0'
call LCDOUT
return

;........

WAVEFRM: incf SHAPE,F
movf SHAPE,W
xorlw %00000011
btfsc STATUS,Z
clrf SHAPE

WAVE1: call LCD1 ;output freq data to LCD
bsf RSLINE,4
clrf LOOPA
WAVE2: movf SHAPE,W
call WAVEIT
call LCDOUT
incf LOOPA,F
btfss LOOPA,2
goto WAVE2

swapf SHAPE,W
movwf STORE
rlf STORE,F
rlf STORE,W
andlw %11000000
movwf STORE
movf PORTB,W
andlw %00111111
iorwf STORE,W
movwf PORTB

WAVE3: btfsc PORTB,0
goto WAVE3
call PAUSIT
return

.END

Need someone to run my mips code please

Noliving — Mon, 02 Nov 2009 22:45:21 GMT

Can someone run my mips code? I'm using pcSpim as my simulator

The issue I'm having is that it doesn't correctly output the number of othercharacters.

Other characters would be like !@#$%^&*()_+-=.,<>?/ \

The other problem I'm having is that it doesn't output the number of characters there are in the string inputted. If I get rid of the othercharacters loop then it will output or display the number of characters or the length of the string inputted.

Thank you to anyone who can help me!

## This program prints out the length of characters, digits, upper and lower case characters that are inputted through the user

## t0 holds each byte from input

## t1 contains the count of characteters

## t2 points to the input of characters



        .data



input:        .asciiz "Please enter up to eighty characters "



ans:        .asciiz "\nThis is how many characters your input has "



anstwo:        .asciiz "\nThis is how many uppercase characters your input has "



ansthree: .asciiz "\nThis is how many lowercase characters your input has "



ansfour: .asciiz "\nThis is how many digits your input has "



ansfive: .asciiz "\nThis is how many spaces your input has "



anssix: .asciiz "\nThis is how many othercharacters your input has "











buffer: .space 80



        .text

        .globl main

main:





        li $v0, 4        # system call code for print_str

        la $a0, input        # address of string to print

        syscall          # print the input



        li $v0, 8        # code for syscall read_string

        la $a0, buffer        # tell syscall where the buffer is

        li $a1, 80        # tell syscall how big the buffer is

        syscall



        la $t2, buffer        # t2 points to the input

        li $t1, 0         # t1 holds the count



        





        la $a0, buffer

        jal         uppercasecount        # call uppercasecount



        la $t2, buffer        # t2 points to the input

        li $t3, 0        # t3 holds the count for how many uppercase



        la $a0, buffer

        jal         lowercasecount        # call lowercasecount



        la $t2, buffer        # t2 points to the input

        li $t4, 0        # t4 holds the count for how many lowercase



        la $t2, buffer        # t2 points to the input

        li $t5, 0        # t5 holds the count for how many digits

        

        jal         digitcount        # call digitcount



        la $t2, buffer        # t2 points to the input

        li $t6, 0        # t6 holds the count for how many spaces

        jal         spacecount        # call spacecount



        



        

        la $t2, buffer

        #li $t7, 0

        jal         Othercharacters



        





        

nextcharacter:





        lb $t0,($t2)        # get a byte from the input

        beqz $t0, characterend        # zero means it is the end of the input

        addi $t1, $t1, 1        # increment the count

        addi $t2,1        # move the pointer to the next character

        j nextcharacter        # go through the next character loop again

        

        



characterend:



        la $a0, ans        # system call to print

        li $v0, 4        # out a message

        sub $t1, $t1, 1        # subtract 1 from count of characters to get correct number

        syscall

        



        move $a0, $t1        # system call to print

        li $v0, 1        # out the length worked out

        syscall



        li $v0,10        # exit

        syscall        



#

#------------------------------------------------

# uppercase - takes a single character as a

# parameter and returns 1 if the character 

# is a uppercase otherwise return 0.

#        a0 - holds character

#        v0 - returns 0 or 1

#------------------------------------------------

uppercase:

        li $v0, 0        #

        beq  $a0,'A',ucyes        #beq means branch is equal to

        beq  $a0,'B',ucyes

        beq  $a0,'C',ucyes

        beq  $a0,'D',ucyes

        beq  $a0,'E',ucyes

        beq  $a0,'F',ucyes

        beq  $a0,'G',ucyes

        beq  $a0,'H',ucyes

        beq  $a0,'I',ucyes

        beq  $a0,'J',ucyes

        beq  $a0,'K',ucyes

        beq  $a0,'L',ucyes

        beq  $a0,'M',ucyes

        beq  $a0,'N',ucyes

        beq  $a0,'O',ucyes

        beq  $a0,'P',ucyes

        beq  $a0,'Q',ucyes

        beq  $a0,'R',ucyes

        beq  $a0,'S',ucyes

        beq  $a0,'T',ucyes

        beq  $a0,'U',ucyes

        beq  $a0,'V',ucyes

        beq  $a0,'W',ucyes

        beq  $a0,'X',ucyes

        beq  $a0,'Y',ucyes

        beq  $a0,'Z',ucyes

        jr $ra         # return (to the operating system)



ucyes:

        li $v0,1        # this is the code perform print_int

        jr $ra        # return (to the operating system)







        jr $ra



#------------------------------------------------

# uppercasecount - use vowelp to count the vowels in a

# string.

#        a0 - holds buffer address

#        t3 - holds number of uppercasecharacters

#        v0 - returns number of uppercasecharacters

#------------------------------------------------



uppercasecount:        

        sub $sp,$sp,16         # save registers on stack

        sw $a0,0($sp)

        sw $t3,4($sp)

        sw $s1,8($sp)

        sw $ra,12($sp)



        li $t3,0        # count of vowels

        move $s1,$a0        # address of string





nextcharacter2:        

        lb $a0,($s1)        # get each character

        beqz $a0,uppercasedone        # zero marks end

        jal uppercase        # call uppercase 

        add $t3,$t3,$v0        # add 0 or 1 to count

        add $s1,$s1,1        # move along string

        j nextcharacter2





uppercasedone:        



        la $a0, anstwo

        li $v0, 4

        syscall



        move $a0, $t3        # system call to print

        li $v0, 1        # out the length worked out

        syscall

        



        lw $a0,0($sp)         # restore registers

        lw $s0,4($sp)

        lw $t3,8($sp)

        lw $ra,12($sp)

        add $sp,$sp,16

        jr $ra





#

#------------------------------------------------

# lowercase - takes a single character as a

# parameter and returns 1 if the character 

# is a uppercase otherwise return 0.

#        a0 - holds character

#        v0 - returns 0 or 1

#------------------------------------------------

lowercase:

        li $v0, 0        #

        beq  $a0,'a',lcyes        #beq means branch is equal to

        beq  $a0,'b',lcyes

        beq  $a0,'c',lcyes

        beq  $a0,'d',lcyes

        beq  $a0,'e',lcyes

        beq  $a0,'f',lcyes

        beq  $a0,'g',lcyes

        beq  $a0,'h',lcyes

        beq  $a0,'i',lcyes

        beq  $a0,'j',lcyes

        beq  $a0,'k',lcyes

        beq  $a0,'l',lcyes

        beq  $a0,'m',lcyes

        beq  $a0,'n',lcyes

        beq  $a0,'o',lcyes

        beq  $a0,'p',lcyes

        beq  $a0,'q',lcyes

        beq  $a0,'r',lcyes

        beq  $a0,'s',lcyes

        beq  $a0,'t',lcyes

        beq  $a0,'u',lcyes

        beq  $a0,'v',lcyes

        beq  $a0,'w',lcyes

        beq  $a0,'x',lcyes

        beq  $a0,'y',lcyes

        beq  $a0,'z',lcyes

        jr $ra         # return (to the operating system)



lcyes:

        li $v0,1        # this is the code perform print_int

        jr $ra        # return (to the operating system)









#------------------------------------------------

# lowercasecount - use lowercase to count the lowercase characters in a

# string.

#        a0 - holds buffer address

#        t4 - holds number of lowercasecharacters

#        v0 - returns number of lowercasecharacters

#------------------------------------------------



lowercasecount:        

        sub $sp,$sp,16         # save registers on stack

        sw $a0,0($sp)

        sw $t4,4($sp)

        sw $s1,8($sp)

        sw $ra,12($sp)



        li $t4,0        # count of lowercase

        move $s1,$a0        # address of string



nextcharacter3:        

        lb $a0,($s1)        # get each character

        beqz $a0,lowercasedone        # zero marks end

        jal lowercase        # call lowercase 

        add $t4,$t4,$v0        # add 0 or 1 to count

        add $s1,$s1,1        # move along string

        j nextcharacter3





lowercasedone:        



        la $a0, ansthree

        li $v0, 4

        syscall



        move $a0, $t4        # system call to print

        li $v0, 1        # out the length worked out

        syscall

        



        lw $a0,0($sp)         # restore registers

        lw $t4,4($sp)

        lw $s1,8($sp)

        lw $ra,12($sp)

        add $sp,$sp,16



        jr $ra         # return (to the operating system)











digit:

        li $v0, 0

        beq  $a0,'1',dyes

        beq  $a0,'2',dyes

        beq  $a0,'3',dyes

        beq  $a0,'4',dyes

        beq  $a0,'5',dyes

        beq  $a0,'6',dyes

        beq  $a0,'7',dyes

        beq  $a0,'8',dyes

        beq  $a0,'9',dyes

        beq  $a0,'0',dyes



        jr $ra         # return (to the operating system)

        



dyes:

        li $v0,1

        jr $ra



digitcount:



        sub $sp, $sp, 16

        sw $a0,0($sp)

        sw $t5,4($sp)

        sw $s1,8($sp)

        sw $ra,12($sp)



        li $t5,0        # count of digits

        move $s1,$a0        # address of string



nextdigit:

        lb $a0,($s1)        # get each character

        beqz $a0,digitdone        # zero marks end

        jal digit        # call digit 

        add $t5,$t5,$v0        # add 0 or 1 to count

        add $s1,$s1,1        # move along string

        j nextdigit



digitdone:        

        la $a0, ansfour

        li $v0, 4

        syscall



        move $a0, $t5        # system call to print

        li $v0, 1        # out the length worked out

        syscall



        lw $a0,0($sp)         # restore registers

        lw $t5,4($sp)

        lw $s1,8($sp)

        lw $ra,12($sp)

        add $sp,$sp,16

        jr $ra



space:

        li $v0, 0

        beq  $a0,' ',syes

        

        jr $ra         # return (to the operating system)



syes:

        li $v0,1

        jr $ra



spacecount:



        sub $sp, $sp, 16

        sw $a0,0($sp)

        sw $t6,4($sp)

        sw $s1,8($sp)

        sw $ra,12($sp)



        li $t6,0        # count of space

        move $s1,$a0        # address of string



nextspace:



        lb $a0,($s1)        # get each character

        beqz $a0,spacedone        # zero marks end

        jal space        # call space 

        add $t6,$t6,$v0        # add 0 or 1 to count

        add $s1,$s1,1        # move along string

        j nextspace



spacedone:

        

        la $a0, ansfive        # how many spaces are in the characters

        li $v0, 4

        syscall



        move $a0, $t6        # system call to print

        li $v0, 1        # out the length worked out

        syscall





        lw $a0,0($sp)         # restore registers

        lw $t6,4($sp)

        lw $s1,8($sp)

        lw $ra,12($sp)

        add $sp,$sp,16

        jr $ra



Othercharacters:

        add $t7, $t3, $t4

        add $t8, $t5, $t6

        add $t9, $t8, $t7

        

        sub $t9, $t2, $t9



        li $v0, 4

        la $a0, anssix

        syscall



        li $v0, 1

        move $a0, $t9

        syscall

Help in using LDR and ADC in 8051

jazz_vill — Mon, 02 Nov 2009 15:49:51 GMT

Hi I need help on how to use LDR and ADC0804 in 8051 microcontroller.
So far I have this in my project but I want to replace the switch with LDR (dark activated).

What happens here (attached file)is that I use the switch to decrement/increment the number of cars and when it reaches to 0(no available space) the LCD will prompt that it is full.

What I want is using a sensor(ldr) and a laser light so that when a car passes (the light will be broken) the ldr will send signal and decrement the number of spaces and vice versa( i'll be using two ldr one for exit the other for entrance).

attached is my design. Im using Isis Proteus 7.4 and the code below

DB0 EQU P1.0



DB1 EQU P1.1



DB2 EQU P1.2



DB3 EQU P1.3



DB4 EQU P1.4



DB5 EQU P1.5



DB6 EQU P1.6



DB7 EQU P1.7







EN EQU P3.7



RS EQU P3.6



RW EQU P3.5



DATA EQU P1







;7-Segment Display



enableDIGIT3 EQU P3.2



enableDIGIT4 EQU P3.3







ZERO    EQU   10001000b



ONE     EQU   10111110b



TWO     EQU   11000100b



THREE   EQU   10010100b



FOUR    EQU   10110010b



FIVE    EQU   10010001b



SIX     EQU   10000001b



SEVEN   EQU   10111100b



EIGHT   EQU   10000000b



NINE    EQU   10010000b



DOT     EQU   01111111b







MOV 20h,#ZERO



MOV 21h,#ONE



MOV 22h,#TWO



MOV 23h,#THREE



MOV 24h,#FOUR



MOV 25h,#FIVE



MOV 26h,#SIX



MOV 27h,#SEVEN



MOV 28h,#EIGHT



MOV 29h,#NINE







MOV R0, #20h



MOV R1, #22h







PDATA EQU P2 ; Connect 8 pins here







CLR 51h        



LCALL LCD_WELCOME



        



Start:



        JB P3.1, check1



                SETB 50h



        check1:



        JB P3.0, check2



                SETB 52h



        check2:



                JNB P3.1, cont1



                        JNB 50h, cont1



                                CLR 50h



                                LCALL INC_7SEGMENT        



        cont1:



                JNB P3.0, cont2



                        JNB 52h, cont2



                                CLR 52h



                                LCALL DEC_7SEGMENT



        cont2:



                LCALL SHOW_7SEGMENT



                LCALL SHOW_LCD



SJMP Start  







SHOW_LCD:



        CJNE @R0, #ZERO, welcome



        CJNE @R1, #ZERO, welcome



                JB 51h, done_showlcd



                LCALL LCD_FULL



        sjmp done_showlcd



        welcome: 



                JNB 51h, done_showlcd



                LCALL LCD_WELCOME



        done_showlcd:



RET



LCD_FULL:        



        LCALL CLEAR_LCD        



        LCALL INIT_LCD



        CLR RS



        MOV DATA,#086H



        SETB EN



        CLR EN



        LCALL WAIT_LCD



        MOV A,#'F'



        LCALL WRITE_TEXT



        MOV A,#'U'



        LCALL WRITE_TEXT



        MOV A,#'L'



        LCALL WRITE_TEXT



        MOV A,#'L'



        LCALL WRITE_TEXT



        MOV A,#'!'



        LCALL WRITE_TEXT



        MOV A,#'!'



        LCALL WRITE_TEXT







        SETB 51h



RET







LCD_WELCOME:



        LCALL CLEAR_LCD        



        LCALL INIT_LCD



        CLR RS



        MOV DATA,#081h



        SETB EN



        CLR EN



        LCALL WAIT_LCD



        



        MOV A,#'P'



        LCALL WRITE_TEXT



        MOV A,#'A'



        LCALL WRITE_TEXT



        MOV A,#'R'



        LCALL WRITE_TEXT



        MOV A,#'K'



        LCALL WRITE_TEXT



        MOV A,#'I'



        LCALL WRITE_TEXT



        MOV A,#'N'



        LCALL WRITE_TEXT



        MOV A,#'G'



        LCALL WRITE_TEXT



        MOV A,#' '



        LCALL WRITE_TEXT



        MOV A,#'S'



        LCALL WRITE_TEXT



        MOV A,#'P'



        LCALL WRITE_TEXT



        MOV A,#'A'



        LCALL WRITE_TEXT



        MOV A,#'C'



        LCALL WRITE_TEXT



        MOV A,#'E'



        LCALL WRITE_TEXT



        



        CLR RS



        MOV DATA,#0C3h



        SETB EN



        CLR EN



        LCALL WAIT_LCD



        



        MOV A,#'A'



        LCALL WRITE_TEXT



        MOV A,#'V'



        LCALL WRITE_TEXT



        MOV A,#'A'



        LCALL WRITE_TEXT



        MOV A,#'I'



        LCALL WRITE_TEXT



        MOV A,#'L'



        LCALL WRITE_TEXT



        MOV A,#'A'



        LCALL WRITE_TEXT



        MOV A,#'B'



        LCALL WRITE_TEXT



        MOV A,#'L'



        LCALL WRITE_TEXT



        MOV A,#'E'



        LCALL WRITE_TEXT



        CLR 51h



RET







INC_7SEGMENT:



        CJNE @R1, #TWO, increment



        sjmp doneinc



        increment:



                CJNE @R0, #NINE, incr0



                MOV R0, #20h



                INC R1



                SJMP doneinc



                incr0:



                        INC R0



        doneinc:



RET







DEC_7SEGMENT:



        CJNE @R0, #ZERO, dec0



        CJNE @R1, #ZERO, dec1



        sjmp donedec



        dec1:



                MOV R0, #29h



                DEC R1



                SJMP donedec



        dec0:



                DEC R0



        donedec:



RET







SHOW_7SEGMENT:



        MOV PDATA,#00h







           setb enableDIGIT3



           MOV     A,@R1



           MOV PDATA,A



           LCALL smallDelay



        clr enableDIGIT3



        



           setb enableDIGIT4



           MOV     A,@R0



           MOV PDATA,A



           LCALL smallDelay           



        clr enableDIGIT4



RET        







INIT_LCD:



        CLR RS



        MOV DATA,#38h



        SETB EN



        CLR EN



        LCALL WAIT_LCD



        



        CLR RS



        MOV DATA,#0Ch



        SETB EN



        CLR EN



        LCALL WAIT_LCD



        



RET







CLEAR_LCD:



        CLR RS



        MOV DATA,#01h



        SETB EN



        CLR EN



        LCALL WAIT_LCD



RET



        



WRITE_TEXT:



        SETB RS



        MOV DATA,A



        SETB EN



        CLR EN



        LCALL WAIT_LCD



RET



        



WAIT_LCD:



        CLR EN ;Start LCD command



        CLR RS ;It's a command



        SETB RW ;It's a read command



        MOV DATA,#0FFh ;Set all pins to FF initially



        SETB EN ;Clock out command to LCD



        LCALL smallDelay



        MOV A,DATA ;Read the return value



        JB ACC.7,WAIT_LCD ;If bit 7 high, LCD still busy



        CLR EN ;Finish the command



        CLR RW ;Turn off RW for future commands



RET  







smallDelay:



                CLR TF1



                CLR TR1



                MOV TH1,#76



                MOV TL1,#01



                MOV TMOD,#01



                SETB TR1



                JNB TF1,$



RET

Attached Images

design.jpeg (170.7 KB)

Max min from array

Koala269 — Mon, 02 Nov 2009 11:17:29 GMT

Write a program in Assembly Language to find the maximum number and the minimum number from an array of ten positive numbers. Store the minimum number in AX and maximum number in DX.

[Hint: Use the conditional jumps]

PLEASE HELP ME!

Need help with a mips program

Noliving — Sat, 31 Oct 2009 20:55:10 GMT

Hello everyone, I'm trying to create a program that will accept a string and output how many characters are in that string. The input can be a mixture of upper/lowercase letters, digits, spaces, and other cases such as .%^& etc.

This is my first time use any type of assembly language so please understand if I have trouble

This is what I have so far, the only thing it does right now is just output what I input, so if I input '"hello" the output will be "hello". Thanks for your help.

        .data



Q1:

        .asciiz "Please enter a string that is up to eighty characters: "



ans1:

        .asciiz "\nThe length of the string you entered contains this many characters: "







buffer:        .space 80        # create space for string input



        .text

        .globl main        #must be global



main:

        li $v0, 4        # system call code for print_str

        la $a0, Q1        # address of string to print

        syscall          # print the Q1

        

        

        li $v0, 8        # code for syscall read_string

        la $a0, buffer        # tell syscall where the buffer is

        li $a1, 80        # tell syscall how big the buffer is

        syscall



        

        li $v0, 8        # code for syscall read_string

        la $t0, buffer        # place address of buffer in $t0

        lb $t1, 0($t0)        # read byte located at address in $t0

        addi $t0, $t0, 0        # increment address in $t0

        #sbrk $a0, 9        # the amount of how long the string is from the address

        



        li $v0, 4

        la $a0, ans1

        syscall

        

        li $v0, 4

        move $a0, $t0

        syscall



        jr $ra        # return to system

add [bx+si],al means what?

kemaletikan — Sat, 31 Oct 2009 20:09:06 GMT

Hi fellas,
I have a question about disassembled code. I have a very simple assembly code that prints "Hello world" to screen. When I disassembled it by using nasm(ndisasm), I got a text file. After that I opened it and started to analyze it. However, in a section that comes after " int 0x80" there are a lot of "add [bx+si],al" statement. What does it means? Can you explain me?

Storing multiple things in a variable?

Goalatio — Fri, 30 Oct 2009 04:47:53 GMT

So, I was trying to re-create the CMD command "set /p", where it prompts the user to type text, and is ended by a carriage return.

Upon trying this, however, I found that a word (dw) will only hold the last thing passed to it. You can type "abcdefg", but the word will only hold the value "g".

What I'm trying to ask is, would there be a way to make it store everything typed, like.. create a string out of a series of characters?

I'm using NASM16 under a windows operating system.

[org 0100h]

%include "Library.asm"



[section .text]



START:

input 16        ;Check if it's enter

je ENDLINE        ;If equal, end the line

mov [store],al        ;Mov AL's value into a variable

string store        ;Display the variable

jmp START



ENDLINE:

string eol        ;End the line

jmp START









[section .data]



store dw 0, "$"

eol db " ", 13, 10, "$"

Help with End of Line character

Goalatio — Thu, 29 Oct 2009 19:27:23 GMT

Hey all, I've been working on this little program in NASM 16 bit assembly, under a windows operating system.

It works, I just have a problem. If you don't pass it any arguments, it fails to see that there is an end of line character.

You pass arguments as..
INKEY a
and it responds "Press a to continue . . ."

If you don't pass anything, by just saying
INKEY
then it responds "Press to continue . . ." and will not exit unless I "End now"

Here's my code, any nudge in the right direction would be appreciated.

NOTE: I am using my own library for this, so if you do not recognize some commands, that is why.. Hopefuly you can see the basic idea of what's happening though, I always try to comment on every line.

[org 0100h]

%include "Library.asm"



[section .text]



string msg1                ;Display "Press"

arg                        ;Grab the arguments off the stack

mov si,bx                ;LODSB!

jmp GRABKEY                ;Get the FIRST argument



GRABKEY:

lodsb                        ;Advance ONE character in SI

cmp al,0d                ;Check if End of Line

je INVALID                ;If it is, exit with ERRORLEVEL 0

mov [character],al        ;Store this character in a word

string character        ;Display the character

string msg2                ;Display "to continue. . ."

jmp INPUTKEY                ;Start asking for the key



INPUTKEY:

input [character]        ;Use library procedure to check\compare AL

                        ;For the key passed as an argument

je VALIDPRESS                ;If it's equal, we can leave!

jmp INPUTKEY                ;Loop if it's not equal



INVALID:

exit 0



VALIDPRESS:

exit 1



[section .data]



character dw 0, "$"

msg1 db "Press ", "$"

msg2 db "to continue . . .", 13, 10, "$"

What does a C++ function header looks like in assembly?

pagis — Thu, 29 Oct 2009 08:19:01 GMT

Hi,

Can anyone please tell me how will a C++ function will start in assembly? What will the header look like? is there a conventional first few rows?

What about the end of a C++ function?

I am very curious about how C++ code really looks like (for example what would a class look like, what would inheritance look like) as I'm trying to understand the "under the hood" issues in c++ opbject oriented code so any referral to a book or an article will be much appreciated.

Thanks!

passing command line parameters mips

shahab03 — Thu, 29 Oct 2009 01:14:49 GMT

I have written a program that multiplies 2 decimals. My multiplication output is fine.

However I would like to do multiplication by passing in two parameters via command line. So far e.g. lets say if I just put '3 4' in the command line without the quotes.. nothing happens...

however if I put the numbers like this 'mult 3 4' -- of course without quote.. its works.. I see 12 on the console... is there anyway where I dont have to type in mult and just type '3 '4' ???

ideas???

Flags used to test conditional jumps...

runtogetdone — Wed, 28 Oct 2009 03:18:21 GMT

Hi, I do not understand how the x86 processor chooses whether to not to jump for a conditional jump. My professor told me that conditional jumps are based on the OF, ZF, SF, and CF flags, but I do not understand which flags are checked under a given jump instruction...

For example, what flags are checked for a unsigned greater-than jump?

Thanks,
Tyler

(In other words, how do I determine the flags in the right column of this website?)
http://siyobik.info/index.php?module=x86&id=146

x86 32-bit registers

brando|away — Wed, 28 Oct 2009 00:03:00 GMT

yeah i can't find this on google but

i understand how to output data using 16bit (with int 21h/ah=02)

say im doing some math calculations, how can i output something in a 32bit register (eax, edx...)?

help with (i think its) floating point arithmetic on 8051

reedpride — Tue, 27 Oct 2009 17:21:04 GMT

; For:                 AT89C4051 @ 11.0592MHz

; Hardware:        INTERFACED WITH ADC0831

; Program goal:        CREATE A DIGITAL VOLTAGE METER

; Assembler:        M-IDE Studio for MCS-51





;place all the usual pre code items here (INTERRUPT VECTORS, ETC)





CS        EQU P3.1                     ;CONNECTED TO CS ON ADC0831

CLK        EQU P3.0                ;CON TO CLK ON ADC0831

D0        EQU P3.7                ;CON TO DO ON ADC0831

;THIS MEANS WE STILL HAVE 1 AVAILABLE PIN LEFT (EXTERNAL INTERRUPT 0, P3.2)



;main

MAIN:        ;INITIALIZE

        ;INITIALIZE

        ;INITIALIZE

        ACALL        ADC0_L

        



;THESE SUBS ARE GOING TO ACTIVATE CHIP, TOGGLE CLOCK, AND GET DIGITAL VALUES STARTED.

ADC0_L:        CLR        CS                ;ACTIVATE CHIP

        MOV        R0,#7                ;8 BIT COUNTER

        SETB        CLK

        ACALL        GET_D0BYTE



GET_D0BYTE:

        CLR        CLK                ;SET LOW

REPEAT:        

        SETB        CLK                ;SET HIGH

        CLR        CLK                ;TOGGLE LOW

        MOV        C,D0                ;MOV TO CARRY

        RRC        A

        DJNZ        R0,REPEAT

        JMP        CALCULATE        ;GO, CALCULATE VALUE

        



CALCULATE:

        MOV        B,0C4H                ;C4 IS 196, WE WANT .0196 BUT THIS WILL DO

        MUL        AB                ;STEP VALUE TIMES WHATEVER OUR CHIP STORED IN ACCUMULATOR

        MOV        R1,B                ;STORE B FOR A FEW

        ;

        ; unfinished

END

Hello, This is my first time posting here, I hope I posted this right. Thanks for taking the time to check this out and potentially help me out.

My goal of this program is to use the Atmel AT89C4051 paired with 7 segment led's (connected to port 1, digit selection for digits 0-4 tied to p1.7/p3.5/p3.4/p3.3 respectively), and help from a ADC0831 chip to create a digital volt meter.

I am only looking for 5 volt max reading with thousandths place accuracy (0.000) to be output to the led display. My led code and wiring is fine (not displayed in code), my only issues are with the math and calculations in order to attain the proper voltage values.

From my understanding of this chip I will activate the ADC0831 chip via chip select (active low) and then toggle the clock while reading pin d0. d0 starts after the second clock low, and continues output on each successive low. I handle this by storing the d0 value each time into carry, rotate carry into the accumulator a total of 8 times. Now the value that d0 gives out is essentially the number that should be multiplied by the count, which from the chip will be 5/255, or a step count of 19.6 mV steps.

******
Lets say the value of d0 is 0A8Hex (168decimal). In assembly Im assuming I want to make the step count value 196 or 0C4Hex and multiply the numbers together to get my voltage value (off by a factor of 4). The true value should be .0196*168decimal=3.292Volts and should be displayed on the LED

now here is my problem. I dont quite understand how to multiply by 196 and get my factor of 4 decimal place error to the true value. I have been stuck on this issue for a while and cannot find any help in my 8051 book. Can anyone please show me or lead me in the right direction for this. If you would like to email me feel free, its just my user name at gmail. Thanks for your time.

Translating into MIPS from C

Woady — Tue, 27 Oct 2009 15:20:24 GMT

I cant figure out how to translate this code into MIPS. Any help or tips would be helpful.

int findMax (int numbers [ ], int count)
{
int currentMax = -1;

if (count > 0)
{
currentMax= 0;
int i = 1;
while (i < count)
{
if (numbers[i] > numbers [currentMax])
currentMax = i;
i = i + 1;
}
}
return currentMax;
}

use $s1 for "currentMax" and $s2 for "i"

Is there a good IDE for TASM?

Arctic wolf — Mon, 26 Oct 2009 18:54:27 GMT

Hello everyone.
I was wondering if anyone could recomend me a good(and bugless) IDE or text editor I could use with TASM(notepad is not very comfortable for code writting)?

Thank you.
P.S. I use TASM 5

TSR Help...

NotNull — Mon, 26 Oct 2009 00:37:01 GMT

I am trying to create an active TSR which prints an 'A'
when the ESC key is pressed, it doesn't work.
Can anyone tell me what's wrong?

bits 16

org 100h



jmp init.tsr



int1c:

push ds

push ax

push cs

pop ds

call checkdos2

jc int1c_e

cmp byte [hotkey_found], 1

jnz int1c_e

call tsrapp

int1c_e:

pop ax

pop ds

iret



int28:

push ds

push ax

push cs

pop ds

call checkdos

jc int28_e

cmp byte [hotkey_found], 1

jnz int28_e

call tsrapp

int28_e:

pop ax

pop ds

iret



checkdos2:

push bx

push es

les bx, [InDOS]

mov al, 0

cmp al, byte [es:bx]

jb checkdos2_e

cmp al, byte [es:bx-1]

checkdos2_e:

pop es

pop bx

ret



checkdos:

push bx

push es

les bx, [InDOS]

mov al, 1

cmp al, byte [es:bx]

jb checkdos_e

xor al, al

cmp al, byte [es:bx-1]

checkdos_e:

pop es

pop bx

ret



InDOS      dd 0

Orig_isr9  dd 0

in_isr9    db 0

hotkey_found db 0

hotkey_idx dw 0

hotseq     db 0x1b



isr9:

push ds

push bx

push ax

push cs

pop ds

in al, 0x60

pushf

cli

call word far [Orig_isr9]

mov ah, [in_isr9]

or ah, [hotkey_found]

jnz isr9_e

inc byte [in_isr9]

cmp al, byte [hotseq]

jz isr9_foundkey

jmp isr9_exit

isr9_foundkey:

mov byte [hotkey_found], 1

isr9_exit:

dec byte [in_isr9]

isr9_e:

pop ax

pop bx

pop ds

iret



tsrapp:

push dx

mov ah, 0x2

mov dl, 0x41

int 0x21

pop dx

ret



pre equ $



init.tsr:

mov ax, 0x3509

int 0x21

mov [Orig_isr9], bx

mov [Orig_isr9+2], es

mov dx, isr9

mov ax, 0x2509

int 0x21

mov dx, int28

mov ax, 0x2528

int 0x21

mov dx, int1c

mov ax, 0x251c

int 0x21

mov ax, 0x3400

int 0x21

mov [InDOS], bx

mov [InDOS+2], es

mov dx, pre

add dx, 0xf

shr dx, 4

mov ax, 0x3100

int 0x21

Pep

G_buchCSC — Sun, 25 Oct 2009 18:11:36 GMT

This is my first time looking at anything in Pep 8 machine language and i was wondering if someone could help me understand how to do these problems . I am supposed to find the output for each program. I know the answer for the first one but I don't know how to get that answer and on the second one i am completely lost.
I've seen 91 is Bitwise AND instruction and as a convert decimal instruction

first problem

0000 C1000E; A: = first number

0003 910010; ?

0006 F1000D; Store the character

0009 51000D; Char output

000C 00; Stop 

000D 00; Character to output

000E A94F; ?

0010 FFFD; ?



Output:

M

Second problem

0000 C1000C; A: = first number

0003 18; ?

0004 F1000B; store the character

0007 51000B; chararacter output

000A 00; Stop

000B 00; Character to outpute

0010 F0D4; ?

Which method is the Correct one , got confused (8085)

namour84 — Sun, 25 Oct 2009 16:02:41 GMT

hello sorry which method is the correct one we solved a question me and my friend in different methods and we went to our prof he said his method is the correct one and i couldn't get his idea so this one the first one is my way of solution
oh first of all the question :P is to write a FUNCTION ((xvy)^z)*2 and where x y and z are memory locations a000h,a001h,a0002h and store the result in a003 and a004 ( 8085)
MY SOLUTION
LXI H, A000H
MOV A,M
INX H
ORA M
INX H
ANA M
MVI B,02H
MOV A,B
LOOP : ADD B
DCR B
JNZ LOOP
STA A003
STA A004
HLT
. THAT WAS MY WAY AND PLZ IF THERE IS A MISTAKE JUST SHOW ME WHERE IS IT AND MY FRIEND SOLUTION WHICH IS THE CORRECT ONE :(

MVI D,00H
LXI H,A00H
MOV A,M
INX H
ORA M
INX H
ANA M
ADD A
JNC LP
INR D
LP : INX H

MOV M,A
INX H
MOV M,D
HLT

THNX IN ADVANCE GUYS

8051 codings for Controlling Movement of the Wheelchair using joystck

shihara — Fri, 23 Oct 2009 10:04:57 GMT

The wheelchair has four wheels, which consist
of two rear wheels and two front wheels. The
two front wheels are pivot wheel which are
set free. There is no actuator which drives
both front wheels. Thus, the front wheels can
move freely in rotation and straight direction.
Therefore, movement of the wheelchair can
only be performed by driving the rear wheels.
Diameter of the front wheel is 10 cm and
diameter of the rear wheel is 22 cm. Two DC
motor are used as the actuator of the
wheelchair. One DC motor drives one rear
wheel.
Specifications of the DC motors are 20 V, 2 A,
and 200 rpm. The DC motor that is used to
actuate rear wheel, has its own gearbox to
reduce the speed of the motor in the ratio of
1:15. Then, speed of the motors is reduced again by using gears and chain system with ratio
1:5. Thus, speed of the motor is totally reduced with ratio 1:75. On the contrary, the torque
of the motor totally increases with ratio 1:75. Because the maximum gear box output speed
of the DC motor is 200 rpm
Joy stick controls the moving direction of the wheelchair (i.e. move forwards or backwards
and turn left or right). Maximum carrying weight of automated wheelchair is 100kg. Once
the weight limit is reached, alarm should be activated and blink a red bulb. Path should be
recorded and store in memory.
There should be a method to recall the path information and come back to the started
location. There should be means to stop the wheelchair whenever an emergency situation is
occurred. Wheelchair should stop if there are any obstacles. Speed of the motors can be
controlled by varying the duty cycle of the PWM (Pulse Width Modulation) signals

code for fan timer

amirnova — Fri, 23 Oct 2009 07:37:42 GMT

hi.is there anyone can help me create a program using esa 8086 trainer kits..??the program need me to turn on the fan@LED for 10seconds and turn off it for 5 seconds..

80x86 Word Memory Access

NotNull — Fri, 23 Oct 2009 03:34:17 GMT

80X86 Speeding up word memory access.
When the address of the first byte of a word lies
at an odd address the processor must make two memory
fetches for each byte of the word.
But if the address of the first byte of the word lies
at an even address the processor can pull in both bytes
of the word in one memory fetch.

If the first word of an array of words lies at an even
address so will every other word in the array.
1st Word__2nd Word__3rd Word
|0100|0101|0102|0103|0104|0105
^________^________^_______
This speeds up performance when accessing an array
of words.

Assemblers usually provide an ALIGN directive to align
a word at an even address.

I was looking through the i386 arch manual on this, and
it mentioned that there could be a slight increase in speed
when the target address of control transfer instructions
were evenly divisible by four, does this apply to the 8086/real-mode?

Good day.

Changing current directory. Help.

Alex_ — Thu, 22 Oct 2009 17:08:13 GMT

Hello everybody. I'm trying to get the current directory, save it, changing the current dir, then coming back to the previous one.

My code so far

TITLE ep1_7

        .MODEL    SMALL

        .STACK    10h

        .DATA

                msg        DB 'Current directory:',0h

                msg_l        equ        $-msg

                buff                db  64 dup('$')

                ;f_handle        dw  1 dup(?)

        .CODE

                begin:   mov        ax,@DATA

                mov        ds,ax

                ;Getting current directory

                mov        ah,47h

                mov        dl,0

                mov        ds,buff

                ;showing the returned pathname

                mov  dx,OFFSET buff

                mov  ah,09h

                int  21h

                ;bail out

                mov        ax,4c00h        

                int        21h        

                END        begin

My problem is that i have an error at line 15. I don't know how to put the buffer's adress into ds. Can some one help me?

Btw , please tell me what is the function that waits for a key to be pressed and/or the one that waits a certain given time to pass.

Question regarding output

RayvenHawk — Thu, 22 Oct 2009 16:33:46 GMT

I have an assignment due for my Assembly class and I have a quick couple of questions regarding it.

The requirement is initialize 3 variables as 16bit words with values (10, -60, 30) add those values as the following equation y = var1 + var2 + var3, use no more than 3 lines of code to produce results. Show results in EAX register.

Question #1: The professor gave us nothing to go off of to ensure we have our code done correctly, so can someone give me the display of the EAX register as it should display if I accomplish the above correctly? (no code, just a screen shot of the output).

Here is the code I have so far:

.data

arrayD WORD 10, -60, 30



.code

mov esi, OFFSET arrayD

add eax, [esi]

add eax, [esi]

add eax, [esi]



call DumpRegs

call WaitMsg



exit

As you can see I have 4 lines of code to produce the results (at least I'm assuming the output is correct). What am I missing that would allow me to drop 1 line of code?

Thanks.

Oh and I realize there is lines missing from the above code, I left them out intentionally since they weren't relevant to my questions.

Beginner SPIM Array help

hwaforum — Thu, 22 Oct 2009 01:15:26 GMT

I have an array

a: .word 0:1000

and I don't understand how I can store/read values from it.

For example if I do

sw $v0, a($s1)

where $s1 is 0 it goes through fine but when I increment $s1 to supposedly access the next space in the array it gives me these errors.

Exception 5 [Address error in store] occurred and ignored
Exception 4 [Address error in inst/data fetch] occurred and ignored

edit: nvm I got it now :P

convert C to Mips

shopnobhumi — Thu, 22 Oct 2009 00:46:39 GMT

I am trying to convert some instruction to MIPS code. But have no idea how to do that.. i know c language so i can decode it to C but dont know how to convert it to MIPS.
1.I need to write a main
2. create an array of 10 integers
3.Declare a variable
4.make a loop to check if it is odd or even
5. print a string saying "it is odd or it is even creating if else statement
6.and increment the variable.
I can write C program for these. But what would it be in MIPS?

int main()

{

int array[10]={1,2,3,4,5,6,7,8,9,10};

int i=0;

while(i<10){

if(i%2==0)

printf("It is even\n");

else 

printf("it is odd\n");

i=i+1;

}

return 0;

}

help with assembly assignment

amw2326 — Wed, 21 Oct 2009 18:06:29 GMT

so I am working on an assignment in class, and I've hit a road block. the assignment is to replace certain letters (b, c) with others, (x, y). Right now, my program is not recognizing 'b' in ebx, and the second letter in the array, 'b', (stored in eax) as being the same, so it is not changing them. I know im doing something wrong, but I can't figure out what. Any hints would be much appreciated, meanwhile i will keep plugging away at it. Also, I cannot get the new array in edx to show up without subtracting sixteen, and im not sure how to fix that so that is not necessary. Thanks so much.



INCLUDE Irvine32.inc

.data

     array byte 'a', 'b', 'c', 'd', 'e', 'f', 'a', 'b', 'c',

                'd', 'e', 'f', 'a', 'b', 'c'

     letter byte ?

     newArray byte 15 dup (?)

.code

main PROC

     mov ecx, lengthof array           ;moves ecx to beginning of array as a counter

     mov esi, offset array           ;moves esi to the beginning of array to obtain letters

     mov eax, [esi]                   ;moves first position of array into eax

    

     doLoop:                        ;starts a loop

     mov ebx, 'b'                   ;moves first letter comparison (b) into ebx

     cmp ebx, eax                    ;compares letters

     je check4C                        ;if it is b, then jump to check for c

     add edx, eax                                        ;if not b, store first letter of newArray in edx

     add esi, type array            ; increment esi to next letter in array



     loop doLoop                    ;loop back to the beginning

   

     mov edx, offset newArray

     sub edx, 16



     call writestring                ;print newArray string with new letters

     call crlf                        ;enters a blank line

     call waitmsg                    ;waits til user closes screen

   

     exit                            ;exit

   

     check4C:                        ;check for c

     add edx, 'x'                ;adds x for b into new array

     add esi, type array            ;move esi down the array

     mov eax, [esi]                    ;move esi into eax for next comparison

     mov ebx, 'c'                    ;move c into ebx for comparison

     cmp ebx, eax                    ;compare registers

     je replace                        ;if the letter is matches c go to replace

     add edx, eax                 ;store next letter of newArray in edx

     add esi, type array            ;move esi down the array

     jmp doLoop                        ;go to the top of the loop

   

     replace:                        ;replaces y for c

     add edx, 'y'                ;moves y into array instead of c

     add esi, type array            ;move esi down the array

     jmp doLoop                        ;go to the top of the loop



main ENDP  



END main

add

shahab03 — Wed, 21 Oct 2009 14:12:32 GMT

I need to write a MIPS program will let the user to enter strings one per line representing valid decimal integers in the range 0 to 999999999999. I need to use 64 bits for each integer. and use some method to convert ascii to integer. here is an example that apparently does it.. but I don't understand it

        addi        $t3, $t1, -48                # Convert from ascii to integer

        mul        $t2, $t2, $t4                # Multiply to put integer in 

                                        # right place

also I am not sure what is meant by using 64 bit integers?

Machine relative offsets

laguna92651 — Wed, 21 Oct 2009 07:57:05 GMT

I am trying to understand how assembly is translated into machine language. I understand how to get the machine coding formats but I'm not sure how to calculate 'rr' in all cases. For the example example bra is 20 rr or 20 03, I know that the relative offset is 03 to get to the iny instruction. But I'm not sure why the rr = F3 for the dbne instruction, how does that relative offset get me to Again? A quick answer would be appreciated, I have a test tomorrow.

Solution to sample problem:

          org $600   

                      ldy  #Array        ;CD B6 15

                      ldaa  #N            ;86 07

                      clr      7,y           ;69 47

Again             brclr   0,y %11110000, Not_gtr  ;0F 40 F0 02

                      bra    Update     ;20 03

Not_gtr          inc     Count       ;72 B6 1C

Update          iny                      ;02

                      dbne  a,Again    ;04 30 F3

                      swi                     ;             

Array             fdb   8,7,2                           

N                   equ   7

Count            rmb   1

Counting 1's in 16 bit word

laguna92651 — Tue, 20 Oct 2009 22:43:15 GMT

I am trying to count the number of 1's in a sixteen bit word for MC68HC12. The only way I can think of is to use 16 BITA instructions, one for each bit location, then increment a counter. I suspect there is a more elegant way to do this. Could someone share their ideas with me.

convert number to ascii in mips?

coni113 — Tue, 20 Oct 2009 20:50:37 GMT

I'm writing a function write_unsigned_int to convert a number to its ascii to write to a string buffer. So far I can only get it to write to the string buffer backwards. Should I create a new buffer and then reverse it after, or is there a simpler way to just write to the buffer in the right order?

.data



buffer: .space  200 

.text



main:

                la                $a1,        buffer

                li                $a2, 231

                li                $a0, 6

                jal                write_unsigned_int

exit:

                addiu              $v0, $0, 4

                la                $a0, buffer

                syscall

                addiu                $v0, $0, 10        

                syscall        





write_unsigned_int:

                # maxlength in $a0, buffer location in $a1, and number to be printed in $a2

                addiu                 $sp, $sp, -8

                sw                 $ra,0($sp)  

                add                $s0, $0, $0 # t0 is the counter

write_loop:

                add                $t2, $0, $a2        # load contents of number

                add                $t3, $s0, $a1        # load buffer



                addiu        $t0, $0, 10

                divu        $a2, $t0

                mfhi        $t0

                addi        $t0, $t0, '0'

                mflo        $a2

                sb                $t0, 0($t3)

                beq                $a2, $0, write_unsigned_int_done

                addi        $s0, $s0, 1

                j                write_loop



write_unsigned_int_done:

                #sb                $a2, 0($t3)

                lw                $ra,0($sp)

                addiu                $sp, $sp, 8

                addu                $v0, $a0, $0

                jr                $ra

thanks in advance!

LOOPZ vs LOOPNZ

assembly101 — Tue, 20 Oct 2009 20:30:37 GMT

I am trying to get this program to use LOOPZ instead of LOOPNZ to find the first positive value in an array but I am not sure how to go about it. Any hints or ideas behind doing it would be much appreciated. This is the code below:

.data
array SWORD -3,-6,-1,-10,10,30,40,4
sentinel SWORD 0

.code
main PROC
mov esi,OFFSET array
mov ecx,LENGTHOF array
next:
test WORD PTR [esi],8000h ; test sign bit
pushfd ; push flags on stack
add esi,TYPE array
popfd ; pop flags from stack
loopnz next ; continue loop
jnz quit ; none found
sub esi,TYPE array ; ESI points to value
quit:

hlt Instruction

dmanw100 — Mon, 19 Oct 2009 21:24:24 GMT

Hello,
I was just wondering if the hlt instruction would produce a blinking cursor at the command line until enter is pressed or if its use is not noticeable to the user. Google wasn't too helpful for the specifics of the instruction. Thanks!

Question about arrays

zmwg — Sun, 18 Oct 2009 23:51:51 GMT

I'm writing a program that takes input for an an array. It then uses a loop to replace the values in another array that has preset values with the values from the array that received input.

When I try to move the values though...well let's say arrayA is the one with the preset values and arrayB is the one that has the inputted values. lw $t3, arrayB($t0) - yields an error, but lw $t3, arrayA($0) works. Is there another command that needs to be used when getting the value from an array that doesn't have preset content?

.data

strOutput: .asciiz "Input: "

arrayA: .word 1,2,3,4

arrayB: .space 16

.text                

        .globl main

main:



        li        $v0, 4

        la        $a0, strOutput

        syscall                



        la         $t1,arrayB

        li        $v0, 8

        la        $a0, 0($t1)

        syscall



        li        $v0, 8

        la        $a0, 4($t1)

        syscall



        li        $v0, 8

        la        $a0, 8($t1)

        syscall



        li        $v0, 8

        la        $a0, 12($t1)

        syscall



        li         $t0, 0

        lw         $t3, arrayB($t0)

I know it's very basic, but I don't know if I'm missing a command or something. Using PCSpim.

And also the number replacement will be done with a loop hence why I'm using arrayB($t0) instead of something else...if that makes any sense at all.

EMU8086 OS Command Comparison

jhouns — Sun, 18 Oct 2009 19:50:23 GMT

Okay, this is an annoying problem i'm making a command line operating system bootable from a floppy disk and the way i'm testing it atm is by inputting a command and outputting it.

The reason for this is that i want to know EXACTLY what its doing with the string and how it is being referenced etc so that i can compare it with a list of commands. The code function 'check' doesnt work at the moment and is very messy what with me playing around with it alot.

Code that would compare the inputted string with a command would be very helpful along with a thorough description OR a description of how i would go about it.

Currently the code is:

;boot code

;==========================================

;Load boot directive

#MAKE_BOOT#



;Set boot memory address

org 7c00h



push cs ;cs = ds

pop ds    

;==========================================

;print welcome

;==========================================      

  

begin:

lea si,welcomeMessage

call print                  



;==========================================

;gets command

;==========================================

mov bh,0

getCommand: 

 lea si,message

 call print

 mov dx,offset buffer

 mov ah,0ah         ;read string subroutine

 int 21h            ;get string

 call check

 jmp getCommand

 

;==========================================

;reboot

;========================================== 

;store 'magic' number at 0040h:0072h

; 0000h = cold reboot

; 1234h = warm reboot

mov ax, 0040h

mov ds,ax

mov w.[0072h],0000h

jmp 0FFFFh:0000h ;reboot!



;==========================================

;procedures

;==========================================



print proc              ;arguments: string is in SI

 nextChar:

  cmp b.[si],0           ;check for null termination

  jz stop

  

  mov al,[si]           ;get next char

  mov ah,0eh            ;teletype subroutine

  int 10h               ;print

  inc si                ;next lesson

  jmp nextChar          ;type next char

 

 stop:   

  ret                   ;return to caller

print endp                              



check proc              ;arguments: string must be in SI

 getNo:

  lea di,help



checkCom:

 xor bx,bx

 lea si,help

 mov bx,offset buffer - 1

 mov ah, 09

temp:

 inc bx

 mov dx,[bx]

 int 21h

 cmp dx,0

 jnz temp

 ret    

check endp

;==========================================

;vars

;==========================================

lfcr equ 13,10,0           



welcomeMessage db "Welcome to MYCROS!"

db 13,10, "For a list of commands type 'help'"

db lfcr



message db 13,10, "Command: ",0



buffer db 50,50 dup " "



help db "help",0

help with MIPS snprintf function?

coni113 — Sun, 18 Oct 2009 18:12:00 GMT

I'm writing a project that's described here:
http://inst.eecs.berkeley.edu/~cs61c...cts/proj2.html

Basically I parse through the *format string and when I see format specifiers like "%x" I insert the correct value that's part of the argument. I'm not sure how to start on this. More specifically, I don't understand how to write to a string buffer.

In the snprintf.s file they provide, they give me a line

buffer: .space  200

so I understand to load the address in say a register $a0, but then what do I do after? Is it a syscall? how would I keep track of where I'm supposed to write to, since I'm passing a string buffer pointer to smaller functions to write the values in.

Also I'm wondering if anyone has ideas on how to parse through the string, would I be reading character by character? (byte by byte) If so, how would I recognize the format specifiers? (%x)

Thanks in advance!

Assembly programming motorola 6800 ebook_example codes please...

assembly — Sun, 18 Oct 2009 15:50:49 GMT

could you help me to find example codes,online adresses,e-books etc. about assembly language for motorola 6800 please....

thank for your helps....

divide overflow problem

urbancalli — Sun, 18 Oct 2009 15:48:46 GMT

MOV AH,DAY

DIV DIVISOR

CMP Al,'0'

JE WEEK_

JMP NOT_WEEK_

that is my code that tries to find out if the "DAY" is divisible by "7", which is the DIVISOR..

i initialized them as..

DAY         DB 1

DIVISOR  DB 7

when the "day" is divisible by 7, a 'divide oveflow' occurs..
how do i get this right? thanks

Displaying contents of registers..?

Ronen444 — Sat, 17 Oct 2009 20:12:25 GMT

Hello, I'm having this little problem, and I will thank you if you help me with it.

What I'm trying to do is to display the contents of various 16-bit registers (e.g. AX, BX, CX, DX), however without success;

Well, what I tried to do is to check the total clusters and total free clusters on the hard-drive using DOS interrupt 21h and function 36h. It returns the number of total clusters into DX.

I tried to print the DX using the function 09h of the same interrupt, however I'm getting gibberish whenever i'm trying to do so.

In short - How can I convert number from registers such as AX or DX into strings, so I can print them to the screen?

why there is ascii code infront?

rayda — Sat, 17 Oct 2009 09:31:22 GMT

i am writing a assembly code to count the number of words in a sentence.
below is my coding:

.model small

.stack 200h

.386

.data



message1 db 10,13, 'Word counter : Enter the sentence to be calculated$'

result db 10,13, 'Number of words : $'

counter db 5,6 dup(0)

sentence db 10,13, 'hello world and me. $' 



.code



Start:

        mov ax,seg message1

        mov ds,ax

        mov dx,offset message1

        mov ah,09h

        int 21h



        mov ax,seg sentence

        mov ds,ax

        mov bx,offset sentence

        mov ax,0

        

        mov cx,0

        mov di,2



check1:

        mov al,[bx+di]



        cmp al,'.'

        je check2

        cmp al,','

        je check2

        cmp al,' '

        je count

        cmp al,'$'

        je done

        jmp skip



check2:

        inc cx

        inc di

        mov al,[bx+di]

        cmp al,' '

        je skip

        jmp check1



skip:

        inc di

        jmp check1



count:

        inc cx

        inc di

        jmp check1



done:

        mov ax,cx

        aaa

        add ax,3030h

        

        cmp ah,30h

        je singleNumber

        jmp doubleNumber



singleNumber:



        mov bx,offset counter

        mov [bx+2],al

        mov al,'$'

        mov [bx+3],al

        jmp print



doubleNumber:

        

        mov bx,offset counter

        mov [bx+2],ah

        mov [bx+3],al

        mov al,'$'

        mov [bx+4],al



print:



        mov dx,offset result

        mov ah,09h

        int 21h



        mov dx,offset counter

        mov ah,09h

        int 21h



.exit

end start

when i run it, there is a ascii symbol infront of the number of word.
isit something wrong with my coding?
please help.
thank you

need help how about assebly error

wilen — Sat, 17 Oct 2009 05:24:41 GMT

i just want to ask how can i add a another value in this code
this is for educational purpose and im not familiar with asm
i know java programming but this is far different

mov eax,[000003E8+100] crash my system

honestly i dont know how to add in asm
any one can help me here

d97aede:

add eax,00

alloc(newmem,256)

label(returnhere)

0D97AC8C:

jmp newmem

nop

returnhere:

newmem:

label(empiesa)

label(setea)

label(sigue)

label(resetea)

alloc(Patiniox,4)

jmp empiesa

empiesa:

cmp [Patiniox],00000000

je setea

jmp sigue

setea:

mov [Patiniox],00000001

jmp empiesa

sigue:

inc [Patiniox]

cmp [Patiniox],00000004

je resetea

jmp d97ac92

resetea:

mov [Patiniox],00000000

jmp d97b2e4

alloc(newmem2,128)

label(returnhere2)

label(originalcode2)

label(exit2)

d97afb5:

jmp newmem2

nop

returnhere2:

newmem2: 

mov eax,000003E8 <-- this part of the code how am i able to add another 100 value here its 1000 so i want to add 100 

mov ecx,000003E8

originalcode2:

add esp,08

mov ecx,[edi+08]

exit2:

jmp returnhere2

Overflow with asr opcode

laguna92651 — Sat, 17 Oct 2009 03:23:11 GMT

I am learning assembly on a MC68HC12.

Can someone explain to me why the overflow flag, V=1, using the asr instruction on following binary number:

%1111 0100
the asr shift would be;
%1111 1010
V=1

another example
%0000 0001
the asr shift would be;
%0000 0000
why does V=1

I think I understand how overflow works with addition and subtraction but haven't been able to find a good explanation for its implementation with the shift instructions. Any recommended reading?
thanks

if statement relative out of range. HELP

masterjiraya — Fri, 16 Oct 2009 12:19:06 GMT

.model small

.stack 64

.data

select db"Enter the your desired option$"

msg1 db "[1]display box and your name$"

msg2 db "[2]enter your name$"

msg3 db "[3]display d to l$"

msg4 db "[4]exit$"

msg5 db "JOSEPHINE P.VINAS$"

.code

mov ax,@data

mov ds,ax



mov ax,0003h

int 10h



mov ah,02

mov dh,0

mov dl,25

int 10h





mov ah, 09h

lea dx, select

int 21h



mov ah,02

mov dh,3

mov dl,5

int 10h



mov ah, 09h

lea dx, msg1

int 21h



mov ah,02

mov dh,4

mov dl,5

int 10h



mov ah, 09h

lea dx, msg2

int 21h



mov ah,02

mov dh,5

mov dl,5

int 10h





mov ah, 09h

lea dx, msg3

int 21h



mov ah,02

mov dh,6

mov dl,5

int 10h



mov ah, 09h

lea dx, msg4

int 21h



mov ah,02

mov dh,1

mov dl,40

int 10h



mov ah,0ah

int 21h



cmp al,'1'

je boxandyourname



cmp al,'2'

je enteryourname



cmp al,'3'

je displayDtoL







jmp exit









;--------------box code------------------------------------…

boxandyourname:

mov ah,02h

mov dh,12

mov dl,30

int 10h





mov ah,09h

mov al,'_'

mov bl,1eh

mov cx,10

int 10h



mov ah,02h

mov dh,17

mov dl,30

int 10h



mov ah,09h

mov al,'_'

mov bl,1eh

mov cx,10

int 10h



mov ah,02h

mov dh,13

mov dl,30

int 10h



mov ah,09h

mov al,'|'

mov bl,1eh

mov cx,1

int 10h



mov ah,02h

mov dh,14

mov dl,30

int 10h



mov ah,09h

mov al,'|'

mov bl,1eh

mov cx,1

int 10h



mov ah,02h

mov dh,15

mov dl,30

int 10h



mov ah,09h

mov al,'|'

mov bl,1eh

mov cx,1

int 10h



mov ah,02h

mov dh,16

mov dl,30

int 10h



mov ah,09h

mov al,'|'

mov bl,1eh

mov cx,1

int 10h



mov ah,02h

mov dh,17

mov dl,30

int 10h



mov ah,09h

mov al,'|'

mov bl,1eh

mov cx,1

int 10h



mov ah,02h

mov dh,13

mov dl,39

int 10h



mov ah,09h

mov al,'|'

mov bl,1eh

mov cx,1

int 10h



mov ah,02h

mov dh,14

mov dl,39

int 10h



mov ah,09h

mov al,'|'

mov bl,1eh

mov cx,1

int 10h



mov ah,02h

mov dh,15

mov dl,39

int 10h



mov ah,09h

mov al,'|'

mov bl,1eh

mov cx,1

int 10h



mov ah,02h

mov dh,16

mov dl,39

int 10h



mov ah,09h

mov al,'|'

mov bl,1eh

mov cx,1

int 10h



mov ah,02h

mov dh,17

mov dl,39

int 10h



mov ah,09h

mov al,'|'

mov bl,1eh

mov cx,1

int 10h





;-------------------name under box-------------------------------------…



mov ah,02

mov dh,20

mov dl,27

int 10h



mov ah, 09h

lea dx, msg5

int 21h

jmp exit

;--------------------enter your name------------------------------------…



enteryourname:

mov ah,02

mov dh,15

mov dl,30

int 10h



mov ah,0ah

int 21h

jmp exit

;----------display d to L----------------------------------

displayDtoL:



mov ah,02

mov dh,10

mov dl,30

int 10h



mov ah,02h

mov cx,9

mov dl,'D'



X:

int 21h

inc dl

loop X





exit:

mov ah, 4ch

int 21h ;end of program

end

why I cannot out put this code it says "relative jump out of range"

Big Issue with Assembly Startup

wiseguy12851 — Fri, 16 Oct 2009 06:33:19 GMT

I've written this Hello World Software with much help in assembly using emu8086 emulator, I'm at chapter 5 of "The Art of Assembly" book and before I continue I need to be able to make this program bootable since I can't very well distribute emulated and virtual software.

I haven't used a floppy drive since the late 90's and I don't plan to use one since no stores around here have any, Virtual floppy drives have never worked in Windows Vista even under administrator mode not including the fact that my program is most likely going to grow over 2 MB.

For over 4 day's I have been searching for a way to get this file onto a CD or just an ISO file for that matter. The emulator / compiler generates a bin (binary) file which I don't know how to copy onto the first sector.

I tried to open is HEX Editor and pad it to 512 bytes with 00 and the boot signature at the end renamed to .bif but Magic ISO could not import so I tried using IMGburn renamed to .img but it needed it to be 1048 bytes or something like that so after padding some more it said it was formatted wrong, there's got to be an easier way.

mips cpu

shahab03 — Thu, 15 Oct 2009 21:05:42 GMT

anyone got any idea how does the CPU determines which register should receive the result produced by an instruction?

thanks

forgot how to repeat a character

masterjiraya — Thu, 15 Oct 2009 14:33:13 GMT

can someone remind me how to loop a character please. I'm begging you all.

my output must be this:

***********************************************
* *
* *
* *
* *
***********************************************

i know how to set the color attribute for a character, my problem is I for got now on how to repeat the character by looping

to mke it proven that It's not a copy paste from some where... I have the color palette here, the cursor position and and the repeatition of the color palette expansion
[CODE]
mov ah, 02
mov dh, 03h
mov dl, 01h
int 10h

mov ah,09h
mov bl,2fh
mov cx, 7
int 10h
[/ICODE]

can some one tell me to remeber what should be added in looping a character

floating point representation

shahab03 — Thu, 15 Oct 2009 13:10:57 GMT

here is an example of floating point representation ... I am unable to understand whats going on. book isnt much help except that I know the bias part and that the matissa is 23 bit and exponent is 8 bits. however I don't understand how the conversion is being done how the 0 and 1 are being generated. can some wise man/woman out there help me understand this... thanks

Convert 0.1015625 to IEEE 32-bit floating point format.
Converting:
0.1015625 × 2 = 0.203125 0 Generate 0 and continue.
0.203125 × 2 = 0.40625 0 Generate 0 and continue.
0.40625 × 2 = 0.8125 0 Generate 0 and continue.
0.8125 × 2 = 1.625 1 Generate 1 and continue with the rest.
0.625 × 2 = 1.25 1 Generate 1 and continue with the rest.
0.25 × 2 = 0.5 0 Generate 0 and continue.
0.5 × 2 = 1.0 1 Generate 1 and nothing remains.
So 0.101562510 = 0.00011012.
Normalize: 0.00011012 = 1.1012 × 2-4.
Mantissa is 10100000000000000000000, exponent is -4 + 127 = 123 = 011110112, sign bit is 0.
So 0.1015625 is 00111101110100000000000000000000 = 3dd0000016

Help please

jedaprick — Thu, 15 Oct 2009 09:15:49 GMT

I HAVE AN .EXE FILE I WANT TO MESS AUROUND WITH
BUT WHEN I OPEN IT WITH NOTEPAD ++ IT LOOKS LIKE THIS GARBAGE CRAP

¦ª2'I¶ÛC„ ˆ¶ÀÕSsÚ4eŒ˜¤ÂÇa1”¹²Ô9&
Û¤éŽ3H ˜©5 †éØM¤•;L˜!S>@÷¤Æ@féRH @ôÆ8Óÿ‡ŠVÊÚ4‚¬(aÝ0&›Cm‹Ïnš´€Ø¦˜Á „l_IË+M4p¥9”¢lsXãCs[CB0É¶aHž|§&ÞPnÛØ

WHAT DO I NEED TO DO TO TRANSFORM IT INTO AN EASY TO READ SCRIPT OR LANGUAGE

DO I ACTUALLY HAVE TO LEARN THIS LANGUAGE TO MESS AUROUND WITH AN EXE FILE QUITE HONESTLY THIS LANGUAGE IS CRAZY PLEASE REPLY ANY ADVICE WOULD HELP
ALL THE BEST ANNONMISS

MIPS Assembly Lanuaguage programming

aks229 — Thu, 15 Oct 2009 05:57:32 GMT

Can anyone please help me with this assignment. Its due friday and i have no clue how to do. thanks in advance for your help.
_____________________________________
Question
Design a MIPS assembly language program to implement the multiplication algorithm of the diagram provided in the link below

http://www.mediafire.com/imageview.p...yzonuq&thumb=5

Demo the program or show screenshots of its execution in an IDE.

need pointers... anyone??

shahab03 — Thu, 15 Oct 2009 04:50:20 GMT

Write a main program for the MIPS machine that uses syscall to repeatedly prompt the user to enter strings one per line representing valid decimal integers in the range 0 to 999999999999. Use 64 bits for each integer. Valid strings may only contain the characters ‘0’ through ‘9’. All strings other than valid decimal integers should be rejected. The numeric value indicated by each of the decimal strings should be added into a running 64-bit sum until a total of 10 strings have been entered or until the user indicates that there is no more input. The computed 64-bit sum should then be displayed on the console.

This exercise involves converting between external and internal representation for input and between internal and external representation for output.

changing interrupt vectors

xixpsychoxix — Thu, 15 Oct 2009 03:56:35 GMT

Hi. I am trying to change an interrupt vector in assembly language to point to a new routine but can't quite figure out how to do it. I am trying to replace it as such:



InterruptRoutine:



  mov ah,0x0

  int 0x16

  iret



InstallInterrupt:



  mov ax,0x0

  mov es,ax

  mov di,Vector*4



  push cs

  pop ax

  stosw

  mov ax,InterruptRoutine

  stosw



  ret



Vector dw 80

This is just an example, I'm actually attempting to change interrupt vector 0x1c, the timer. I am writing this as standalone code as i am attempting to write a mini-os, and this code will run at system boot so i cant use dos to do this for me. is this the correct way of doing this? If so, why isn't the timer firing under my new code? I have no clue what is going on here. And is this the right way to find the segment for the routine or is there a better way?

Could anyone please tell me?

punchinello — Thu, 15 Oct 2009 03:33:38 GMT

What is a segment selector?
How is "overflow" defined in assembly programming language?

Thank you very much in advance!

can someone tell if this output is correct?

shahab03 — Wed, 14 Oct 2009 22:41:19 GMT

so i have this following code which does this simple thing in mips

mult $t0, $t1

$t0 = -763
$t1 = 29

when I move the values from hi and lo registers to $t2 and $t3 i get following values for each
$t2 = -1 and $t3 = -22127

is this correct??? can someone explain me why?

here is the code



        .text 

        .globl main

main:



addi $t0,$zero,-763

move $a0,$t0

li $v0, 1

syscall



la $a0, plus

li $v0, 4        #print_string

syscall



addi $t1,$zero,29

move $a0,$t1

li $v0, 1

syscall





mult        $t0,$t1



mfhi $t2

mflo $t3



la $a0, t0equals

li $v0, 4        #print_string

syscall



move $a0,$t0

li $v0, 1

syscall





la $a0, t1equals

li $v0, 4        #print_string

syscall





move $a0,$t1

li $v0, 1

syscall



la $a0, t2equals

li $v0, 4        #print_string

syscall

        

move $a0,$t2

li $v0, 1

syscall



la $a0, t3equals

li $v0, 4        #print_string

syscall



move $a0,$t3

li $v0, 1

syscall



li $v0, 10 # syscall code 10 is for exit.

syscall 



        .data

prompt:         .asciiz "Please enter a (floating point) number: "

plus:         .asciiz " + "

t0equals:         .asciiz "\n t0 = "

t1equals:         .asciiz "\n t1 = "

t3equals:         .asciiz "\n t3 = "

t2equals:         .asciiz "\n t2 = "

# end of add2.asm.

Looking for good Assembler/C++ programmer

Jony23 — Wed, 14 Oct 2009 20:17:36 GMT

Well i am new in this forum so i am not sure which section this topic should be in.

I am running global server of mmorpg, and although the new "fixes" we tried and MHS blocking, they could still find a way to bypass it.

therefor i need someone who knows well in this stuff to prevent it , and to make an unhackable client.exe/login.exe for our server.

he can do it for fun or for money, as he want.

Please contact me Via messenger.

digobest123@hotmail.com

or send me PM

Thanks.

job

tahiralikhan — Wed, 14 Oct 2009 13:39:39 GMT

tahir ali

Command-line arguments?

Goalatio — Wed, 14 Oct 2009 10:55:17 GMT

Would anyone know how to get command-line arguments in NASM16? I am on a Windows XP. I have tried a great deal of things and spent much time searching for an answer to this.

What I ment by command line arguments, something simple like this would be enough.

Filename: Hello1
Params: one two three

So when you type "hello1 one two three", it would see that you typed "one two three", and say something along the lines of

Param 1: One
Param 2: Two
Param 3: Three

I am really against using C libraries (I prefer to do everything purely in assembly).. So if it is possible to do it without C, I would really appreciate your time to answer.

Someone may ask "Why would you want to do this? Just use a higher level language.".. My short answer would be that I enjoy using assembly over a high level language, because of its complexity.

C equivalnet Assembly

Gaiety — Wed, 14 Oct 2009 04:39:13 GMT

Hi,

i am presently working on C but there are some situations i need to know the assembly code of corresponding C program.

if any can provide some general info of c - assembly it would be appreciated.

i have a program called

int main()

{

       printf("Hello,Plz Help\n");

       return 0;

}

if i run this program on Gcc compiler with -S option i could see some assembly instructions.

this is what it is generating . but i am unable to understand
it.below are few questions.

.file   "assm-c.c"

        .section        .rodata

.LC0:

        .string "Hello, Plz help\n"

        .text

.globl main

        .type   main,@function

main:

        pushl   %ebp

        movl    %esp, %ebp

        subl    $8, %esp

        andl    $-16, %esp

        movl    $0, %eax

        subl    %eax, %esp

        subl    $12, %esp

        pushl   $.LC0

        call    printf

        addl    $16, %esp

        movl    $0, %eax

        leave

        ret

.Lfe1:

        .size   main,.Lfe1-main

        .ident  "GCC: (GNU) 3.2.2 20030222 (Red Hat Linux 3.2.2-5)"

what is .rodata
what is .LC0
and what is the procedure that is written in main:
it may be internal processor dependent what it is generating here.
but i want to know what are the general things that are done when converting to assembly.

i have some idea with fucntions
when ever a function is called .
1.push the arguments from right to left into the stack ( meaning allocate memory on stack)
2.push the return address of the function(caller's address)
3.push the base address of functions(caller' s again)
4. allocate memory for the locals.

and when the function returns clering the contents from the stack(dont have idea what does this mean)

is the above procedure correct or do i need to make any changes.

requesting you to guide me in correcting my mistakes and furthur updations.

Auxiliary Carry Flag Help

assembly101 — Wed, 14 Oct 2009 01:04:40 GMT

I am trying to find out if the Auxiliary carry flag only turns on when the 4th bit carries out from the 3rd bit or if it will turn on for the 7th to 8th bit aswell. This is what I did already:

mov ax, 0Fh ; this is the test for the 3rd to 4th
add ax, 1 ; AC=1

mov ax, 00F0h ; this is the test for the 7th to 8th
add ax, 0010h ; not sure if this should set AC=1

total rookie,pls help me

helinix — Tue, 13 Oct 2009 15:02:05 GMT

Hi , i want to write Mips Assembly code to increment the elements of an array. i've written this :
#AIncrement.s -- A program that increments the elements of an array
##################################
main:
.data

list : .word 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
.text
move $t4,$zero
addi $t4,$t4,1
addi $t4,$t4,1
addi $t4,$t4,1
addi $t4,$t4,1
Loop:

la $s0, list # put address of list into $s0
move $t3,$zero # loop variable= 0 int i = 0
add $t1,$t3,$t3 #Temp reg $t1 = 2 * i
add $t1,$t1,$t1 #Temp reg $t1 = 4 * i
add $t1,$t1,$s0 #$t1 = address of A[i]
lw $t0,0($s0) #Temporary reg $t0 = A[i]
addi $t0,$t0,1 #incrementing
sw $t0,0($s0)
addi $t3,$t3,1
bne $t3,$t4,Loop

li $v0,10
syscall
#end of the program
########################
When i use SPIM simulator the file is successfully loaded, but when i
hit the go button just nothing happens, is there anything wrong with my code ?

thanks

Array input question in assembly

heat89 — Mon, 12 Oct 2009 21:12:57 GMT

I would like to be able to have the user input a 1 or 0 and have whatever was inputted into an array. Here is what I have so far.

.model small

.stack 100h

.code

start:
my_arr db ?
mov dl, [my_arr]
mov [my_arr], al
add dl, '0'
mov ah, 2h
int 21h

end start

Basically I'm trying to get it to input into the array, then testing if the it worked by outputting the first element in the array. Right now all it outputs is N..which I'm assuming stands for null meaning nothing is being inputted in the array based on this code. Can somebody help me out?

floating point syscall

shahab03 — Mon, 12 Oct 2009 18:59:33 GMT

what is READ_FLT? any thoughts on the question?

If a program employs a READ_FLT syscall to input the floating point value 999999999999.0, by how much would the value returned by the syscall differ from the actual value typed in? Explain the error, if any.

Assembly inside a C++ program

ubuntuubuntu — Mon, 12 Oct 2009 17:17:25 GMT

Hello,
I have a program, let's say foo.cpp . If I do "g++ -S foo.cpp". That generates the file "foo.S" with assembly code. I would like to compare the assembly code foo.S with the assembly code generated if I put foo.S in a c program (even though there should be no difference), but I don't know how to do it. For foo.cpp =

#include 



using namespace std;

int main() {

  cout << "a47" << endl;

  return 0;

}

for example, I get foo.S =

        .mod_init_func

        .align 2

        .long        __GLOBAL__I_main

.lcomm __ZSt8__ioinit,1,0

        .cstring

LC0:

        .ascii "a\0"

        .text

        .align 1,0x90

.globl _main

_main:

LFB1480:

        pushl        %ebp

LCFI0:

        movl        %esp, %ebp

LCFI1:

        pushl        %ebx

LCFI2:

        subl        $20, %esp

LCFI3:

        call        L3

"L00000000001$pb":

L3:

        popl        %ebx

        leal        LC0-"L00000000001$pb"(%ebx), %eax

        movl        %eax, 4(%esp)

        leal        L__ZSt4cout$non_lazy_ptr-"L00000000001$pb"(%ebx), %eax

        movl        (%eax), %eax

        movl        %eax, (%esp)

        call        L__ZStlsISt11char_traitsIcEERSt13basic_ostreamIcT_ES5_PKc$stub

        movl        %eax, %edx

        leal        L__ZSt4endlIcSt11char_traitsIcEERSt13basic_ostreamIT_T0_ES6_$non_lazy_ptr-"L00000000001$pb"(%ebx), %eax

        movl        (%eax), %eax

        movl        %eax, 4(%esp)

        movl        %edx, (%esp)

        call        L__ZNSolsEPFRSoS_E$stub

        movl        $0, %eax

        addl        $20, %esp

        popl        %ebx

        leave

        ret

LFE1480:

        .align 1,0x90

___tcf_0:

LFB1489:

        pushl        %ebp

LCFI4:

        movl        %esp, %ebp

LCFI5:

        pushl        %ebx

LCFI6:

        subl        $20, %esp

LCFI7:

        call        L6

"L00000000002$pb":

L6:

        popl        %ebx

        leal        __ZSt8__ioinit-"L00000000002$pb"(%ebx), %eax

        movl        %eax, (%esp)

        call        L__ZNSt8ios_base4InitD1Ev$stub

        addl        $20, %esp

        popl        %ebx

        leave

        ret

LFE1489:

        .section __TEXT,__StaticInit,regular,pure_instructions

        .align 1

__Z41__static_initialization_and_destruction_0ii:

LFB1488:

        pushl        %ebp

LCFI8:

        movl        %esp, %ebp

LCFI9:

        pushl        %ebx

LCFI10:

        subl        $36, %esp

LCFI11:

        call        L12

"L00000000003$pb":

L12:

        popl        %ebx

        movl        %eax, -12(%ebp)

        movl        %edx, -16(%ebp)

        cmpl        $65535, -16(%ebp)

        jne        L11

        cmpl        $1, -12(%ebp)

        jne        L11

        leal        __ZSt8__ioinit-"L00000000003$pb"(%ebx), %eax

        movl        %eax, (%esp)

        call        L__ZNSt8ios_base4InitC1Ev$stub

        leal        L___dso_handle$non_lazy_ptr-"L00000000003$pb"(%ebx), %eax

        movl        (%eax), %eax

        movl        %eax, 8(%esp)

        movl        $0, 4(%esp)

        leal        ___tcf_0-"L00000000003$pb"(%ebx), %eax

        movl        %eax, (%esp)

        call        L___cxa_atexit$stub

L11:

        addl        $36, %esp

        popl        %ebx

        leave

        ret

LFE1488:

        .align 1

__GLOBAL__I_main:

LFB1490:

        pushl        %ebp

LCFI12:

        movl        %esp, %ebp

LCFI13:

        subl        $8, %esp

LCFI14:

        movl        $65535, %edx

        movl        $1, %eax

        call        __Z41__static_initialization_and_destruction_0ii

        leave

        ret

LFE1490:

        .section __TEXT,__eh_frame,coalesced,no_toc+strip_static_syms+live_support

EH_frame1:

        .set L$set$0,LECIE1-LSCIE1

        .long L$set$0

LSCIE1:

        .long        0x0

        .byte        0x1

        .ascii "zPR\0"

        .byte        0x1

        .byte        0x7c

        .byte        0x8

        .byte        0x6

        .byte        0x9b

        .long        L___gxx_personality_v0$non_lazy_ptr-.

        .byte        0x10

        .byte        0xc

        .byte        0x5

        .byte        0x4

        .byte        0x88

        .byte        0x1

        .align 2

LECIE1:

        .globl _main.eh

_main.eh:

LSFDE1:

        .set L$set$1,LEFDE1-LASFDE1

        .long L$set$1

LASFDE1:

        .long        LASFDE1-EH_frame1

        .long        LFB1480-.

        .set L$set$2,LFE1480-LFB1480

        .long L$set$2

        .byte        0x0

        .byte        0x4

        .set L$set$3,LCFI0-LFB1480

        .long L$set$3

        .byte        0xe

        .byte        0x8

        .byte        0x84

        .byte        0x2

        .byte        0x4

        .set L$set$4,LCFI1-LCFI0

        .long L$set$4

        .byte        0xd

        .byte        0x4

        .byte        0x4

        .set L$set$5,LCFI3-LCFI1

        .long L$set$5

        .byte        0x83

        .byte        0x3

        .align 2

LEFDE1:

___tcf_0.eh:

LSFDE3:

        .set L$set$6,LEFDE3-LASFDE3

        .long L$set$6

LASFDE3:

        .long        LASFDE3-EH_frame1

        .long        LFB1489-.

        .set L$set$7,LFE1489-LFB1489

        .long L$set$7

        .byte        0x0

        .byte        0x4

        .set L$set$8,LCFI4-LFB1489

        .long L$set$8

        .byte        0xe

        .byte        0x8

        .byte        0x84

        .byte        0x2

        .byte        0x4

        .set L$set$9,LCFI5-LCFI4

        .long L$set$9

        .byte        0xd

        .byte        0x4

        .byte        0x4

        .set L$set$10,LCFI7-LCFI5

        .long L$set$10

        .byte        0x83

        .byte        0x3

        .align 2

LEFDE3:

__Z41__static_initialization_and_destruction_0ii.eh:

LSFDE5:

        .set L$set$11,LEFDE5-LASFDE5

        .long L$set$11

LASFDE5:

        .long        LASFDE5-EH_frame1

        .long        LFB1488-.

        .set L$set$12,LFE1488-LFB1488

        .long L$set$12

        .byte        0x0

        .byte        0x4

        .set L$set$13,LCFI8-LFB1488

        .long L$set$13

        .byte        0xe

        .byte        0x8

        .byte        0x84

        .byte        0x2

        .byte        0x4

        .set L$set$14,LCFI9-LCFI8

        .long L$set$14

        .byte        0xd

        .byte        0x4

        .byte        0x4

        .set L$set$15,LCFI11-LCFI9

        .long L$set$15

        .byte        0x83

        .byte        0x3

        .align 2

LEFDE5:

__GLOBAL__I_main.eh:

LSFDE7:

        .set L$set$16,LEFDE7-LASFDE7

        .long L$set$16

LASFDE7:

        .long        LASFDE7-EH_frame1

        .long        LFB1490-.

        .set L$set$17,LFE1490-LFB1490

        .long L$set$17

        .byte        0x0

        .byte        0x4

        .set L$set$18,LCFI12-LFB1490

        .long L$set$18

        .byte        0xe

        .byte        0x8

        .byte        0x84

        .byte        0x2

        .byte        0x4

        .set L$set$19,LCFI13-LCFI12

        .long L$set$19

        .byte        0xd

        .byte        0x4

        .align 2

LEFDE7:

        .section __IMPORT,__jump_table,symbol_stubs,self_modifying_code+pure_instructions,5

L___cxa_atexit$stub:

        .indirect_symbol ___cxa_atexit

        hlt ; hlt ; hlt ; hlt ; hlt

        .section __IMPORT,__pointers,non_lazy_symbol_pointers

L__ZSt4cout$non_lazy_ptr:

        .indirect_symbol __ZSt4cout

        .long        0

        .section __IMPORT,__jump_table,symbol_stubs,self_modifying_code+pure_instructions,5

L__ZStlsISt11char_traitsIcEERSt13basic_ostreamIcT_ES5_PKc$stub:

        .indirect_symbol __ZStlsISt11char_traitsIcEERSt13basic_ostreamIcT_ES5_PKc

        hlt ; hlt ; hlt ; hlt ; hlt

L__ZNSolsEPFRSoS_E$stub:

        .indirect_symbol __ZNSolsEPFRSoS_E

        hlt ; hlt ; hlt ; hlt ; hlt

        .section __IMPORT,__pointers,non_lazy_symbol_pointers

L__ZSt4endlIcSt11char_traitsIcEERSt13basic_ostreamIT_T0_ES6_$non_lazy_ptr:

        .indirect_symbol __ZSt4endlIcSt11char_traitsIcEERSt13basic_ostreamIT_T0_ES6_

        .long        0

L___dso_handle$non_lazy_ptr:

        .indirect_symbol ___dso_handle

        .long        0

        .section __IMPORT,__jump_table,symbol_stubs,self_modifying_code+pure_instructions,5

L__ZNSt8ios_base4InitD1Ev$stub:

        .indirect_symbol __ZNSt8ios_base4InitD1Ev

        hlt ; hlt ; hlt ; hlt ; hlt

        .section __IMPORT,__pointers,non_lazy_symbol_pointers

L___gxx_personality_v0$non_lazy_ptr:

        .indirect_symbol ___gxx_personality_v0

        .long        0

        .section __IMPORT,__jump_table,symbol_stubs,self_modifying_code+pure_instructions,5

L__ZNSt8ios_base4InitC1Ev$stub:

        .indirect_symbol __ZNSt8ios_base4InitC1Ev

        hlt ; hlt ; hlt ; hlt ; hlt

        .constructor

        .destructor

        .align 1

        .subsections_via_symbols

I thought this would work, but it doesn't:

asm(" ** ALL THE ASSEMBLY CODE ABOVE ** ");

A friend suggested me to use
perl -pi~ -e ’s/\"/\\\"/g; s/(.*)/\"$1\\\n\"/ig’ foo.s
but it didn't work either.
Does anyone know how to do this? Thank you in advance!

help with question

shahab03 — Mon, 12 Oct 2009 14:09:10 GMT

hi.. can someone help me with this question

Write down a sequence of instructions that would implement the synthetic instruction addd $t0,$t2 to produce a double precision sum (i.e. 64-bit sum) of two unsigned 64-bit integers. The first 64-bit integer is in the register pair $t0 and $t1, while the second 64-bit integer is in the register pair $t2 and $t3 (the even numbered register in each pair contains the most significant bits). The result should be left in the $t0,$t1 register pair. Your solution should not modify any registers other than $t0 and $t1 ( and $1 if needed). Nor should your solution involve any type of branch or jump instruction.

I understand that synthetic instructions are like

add $t1,$t0,$a1

since the question is asking for sum of two unsigned 64 bit integers.. what does that mean? aren't unsigned integers 32 bit?
thanks

Changing Text Colour with NASM Assembly

MichaelSammels — Mon, 12 Oct 2009 10:55:15 GMT

Does anyone know how to change the colour of strings using x86 NASM Assembler? I've found tutorials like this:

mov  ah,  09h 

mov  al,  97 

mov  bx,  100b

mov  cx,  01h 

int  10h

Which prints the letter "A" once in the colour you choose. This is a handy tutorial, but I would like to be able to print the entire screen text (i.e more than one string) in the same colour. Anyone have any ideas?

CPUID :)

chrisname — Sun, 11 Oct 2009 12:51:44 GMT

Hi :)

I'm trying to get the vendor ID using the CPUID opcode. Again, this is inline ASM with C; but it's the ASM that I need help with. The C is fine AFAIK.

I'm not sure if it's the way I'm trying to do this, or if my CPU simply doesn't support CPUID...

char* getCPUID(void) {

    int vendor[3];

    char vstring[12 + 1];



    /* Get vendor string: */

    asm("mov %0, %%eax\n\t"

        "cpuid\n\t"

        "cmp $3, %%eax\n\t" /* If eax < 3, CPU doesn't support CPUID... */

        "jl dont_bother\n\t" /* So we can't CPUID :( */

        "mov %%ebx, %0\n\t"

        "mov %%ecx, %1\n\t"

        "mov %%edx, %2\n\t"

        "dont_bother:\n\t"

        "mov $-1, %0"

        :"=r"(vendor[0]), "=r"(vendor[1]), "=r"(vendor[2])

       );

       

   if (vendor[0] == -1) {

       return "Error: CPU doesn't support CPUID!";

   }



   /* Store in vstring; 12 char string results */

   memcpy(&(vstring[0]), &(vendor[0]), sizeof(int));

   memcpy(&(vstring[4]), &(vendor[1]), sizeof(int));

   memcpy(&(vstring[8]), &(vendor[2]), sizeof(int));



   /* Make sure string is NULL-terminated */

   vstring[12] = '\0';

   

   return getManufacturer(vstring); /* Function simply compares vstring to some

                                       known vendor strings. */

}

... but my error message is printed. Is it my ASM, or the CPU?
I'm hoping it's the ASM...

Quote:

$ ./a.out
Manufacturer: Error: CPU doesn't support CPUID!

NASM Painting Pixels

tomtetlaw — Sat, 10 Oct 2009 06:05:37 GMT

How do you paint pixels in NASM, could someone tell me about it and mabey give me an example?

Countdown timer

nikolai090 — Fri, 09 Oct 2009 18:16:07 GMT

Hi there, I'm kinda new to making asm programs and i was wondering if you could make a countdown timer where the user is asked for the input and it is displayed in countdown, lets say a max of 59 secs. I was only able to make one that counts down from 9 to 0, is there a way to input 59secs and display the countdown? Using A86 assembler btw.

Here's my code:

MAIN:

                JMP START

                MSG: DB 0A,0D,'ENTER NUMBER OF SECONDS(0-9): $'

                MSG1: DB 0A,0D,'-END- $'



START:        ;USER INPUT

                MOV AH,09

                MOV DX,MSG

                INT 21H

SECIN:

                MOV AH,08

                INT 21H

                PUSH AX

                MOV AH,02

                MOV DL,AL

                INT 21H

                MOV AH,02

                MOV DL,0AH

                INT 21H

                POP AX

                MOV BL,AL

                MOV CL,AL

DCOUNT:

                CMP BL,30H

                JE COUNTEND

                PUSH CX

                PUSH BX

                MOV AH,02

                MOV DL,BL

                INT 21H

                POP BX

                DEC BL

                MOV AH,02

                MOV DL,0AH

                INT 21H

                CALL DELAYT

                POP CX

                LOOP DCOUNT

                INT 20H

                

DELAYT:        ;DELAY FUNCTION USING SERVICE 2C OF INT 21H

                MOV AH,02CH

                INT 21H

                MOV BH,DH  ; DH has current second

TIMEC:      ; Loops until the current second is not equal to the last, in BH

                MOV AH,02CH

                INT 21H

                CMP BH,DH  ; Here is the comparison to exit the loop continue

                JE TIMEC

                RET

                

COUNTEND:

                MOV AH,09

                MOV DX,MSG1

                INT 21H

                INT 20H

jl and jg?

jingo1126 — Fri, 09 Oct 2009 14:08:17 GMT

ok i have this problem in mind i wanted to try but i cant figure out whats wrong with my code..the program lets you input a string of integers then press another input as target..The target is then cmp to the string..the output should place lesser values on the string based on the specifed target ex.
string: 12456
target: 2
output: 1

.model small

.stack

.data

.code

start:

push 0ah ; push linefeed indication of stack

start1:

mov ah,1

int 21h

push ax      ; data place at stack

cmp al,0dh   

je out1       ; will jump if equal to cmp al,0dh

jne start1 ; will jump if not equal to cmp al,0dh

out1:

mov ah,2

mov dl,0ah ;linefeed

int 21h



target:

pop bx

mov ah,1

int 21h

tests:

cmp al,bl

jl out2

out2:

pop bx

mov ah,2

mov dl,al

int 21h

cmp al,0ah

jne tests

je fin

fin:

mov ah,4ch

int 21h

end start

need help...im just wondering how come it wont workout the way i wanted it to workout..

Question

evant8950 — Fri, 09 Oct 2009 01:42:54 GMT

I need to write a program in assembly that will read a file using the irvine library. My question is if I read in 50 bytes into a buffer, I can't figure out how to get the second 50 bytes and so on? Does anyone have an idea on how to do this.

Thanks

the address offset's time of system

c+++ — Wed, 07 Oct 2009 11:15:19 GMT

hi, who can give me the address offset's time of system

my first program

tomtetlaw — Wed, 07 Oct 2009 08:12:00 GMT

I am trying to make a program that prompts the user for input then outputs it to them(back to the good old days :D). It all works, except it doesn't output what the user inputs, it outputs a bunch of crazy numbers, can someone help me?

Btw I am using NASM.

Test Run:

Please enter something(max 255 chars): tom

You entered: 7

[section .data]

        choice:  db ""

        fmt1:    db "%s",10,0

        output:  db "Please enter something(max 255 chars): ",0

        output2: db "You entered: %s",0

[section .text]

        global _main 

        extern _printf, _scanf 

_main:

        push output

        call _printf

        add esp, 4

        

        push choice

        push fmt1

        call _scanf

        add esp, 8

        

        push dword[choice]

        push output2

        call _printf

        add esp, 8

        

        mov eax, 0

        ret

MIPS Checking if Integer/Letter

gauntex — Wed, 07 Oct 2009 04:31:10 GMT

I'm having trouble with my program. I believe I've figured out most of it, but there's one thing that I still have no clue about. That's how to go through a string and check whether it's a digit or letter.

I figured I could do something like this to remove whitespace (although I haven't tested it out yet... just saw this while searching other topics here):

li $t0, 32             # store a whitespace into temp variable

bne $t1, $t0, remove     # if not a whitespace branch to remove

Assuming $t1 is a character I want to check (and remove is defined later in the code).

But that doesn't handle symbols or anything. I'm thinking it may be easier to just check if the character IS a digit/letter instead of checking if it isn't. But I don't know how to do that either.

One more thing. Is there any way to take a input file as a argument? I realize I can prompt the user for input like this:

                  .data

prompt:          .asciiz "Input: "

buf:             .space 100

                  .text

main:

li                 $v0, 4                           # system call code for print_string

la                 $a0, prompt

syscall



li                 $v0, 8                            # system call code for read_string

la                 $a0, buf

li                 $a1, 100

syscall

But can I actually read input from a file given as a argument?

initialize a serial port in protected mode

therstonsplace — Wed, 07 Oct 2009 02:48:37 GMT

i have been trying to do any interrupts in protected mode and i have been using msdos interrupts so it makes sense that they are not working i have been working through the threads so as not duplicate this but i haven't found any and i figured i would just ask

im kind of new to assembly but im doing alright so far

openfile hasn't worked and just calling int 21h or int 14h hasn't worked

in fact if i just call sti to prepare the assembler for interrupts i get a problem not to mention trying actuall interrupts

i need to open a serial port in assembly using protected mode....

as well as set the baud rate

any ideas

thanks ahead of time and whatever i need to read i will...

thom

mips recursive

shahab03 — Tue, 06 Oct 2009 22:59:55 GMT

I am trying to do towers of hanoi recursively below is the code. my problem is that i am not getting out of the cycle at the right time so code keeps on running... can someone tell me what I may be doing wrong?

.data

msg0: .asciiz "Enter number of disks:"

msg1: .asciiz "Move disk 1 from TOWER "

msg2: .asciiz "Move disk "

msg3: .asciiz " from TOWER "

msg4: .asciiz " to TOWER "

msg5: .asciiz "\n"



.text

.globl main

main:        





la $a0,msg0        #Print "Enter number of disks:"

li $v0,4

syscall



li $v0,5         #cin, $v0=input value

syscall



addi $t0,$zero,6        #$t0=6

move $a2,$v0        #$a2=number of disc,initialized by input value

addi $t7,$zero,1        #$t7=from,initialize to 1

addi $a1,$zero,3        #$a1=to,initialize to 3

move $t1,$t0        #$t1=temp,initialize to $t1=6



addi $sp,$sp,-20         #create a stack with 5 boxes

sw $ra,16($sp)         #$ra |retun address|

sw $t7,12($sp)         #$t7 |from |

sw $a1,8($sp)         #$a1 |to |

sw $a2,4($sp)         #$a2 |numberof disc|

sw $t1,0($sp)         #$t1 |temp |<---$sp



jal hanoi # call hanoi



#lw $ra,16($sp)         # -----------------(*)



lw $ra,16($sp)         #$ra |retun address|

lw $t7,12($sp)         #$t7 |from |

lw $a1,8($sp)         #$a1 |to |

lw $a2,4($sp)         #$a2 |numberof disc|

lw $t1,0($sp)         #$t1 |temp |<---$sp



addi $sp,$sp,20

#jr $ra





hanoi:





sub $t1,$t0,$t7

sub $t1,$t1,$a1         #temp = 6-from-to







addi $sp,$sp,-20         #create a stack with 5 boxes



sw $ra,16($sp)         #$ra |retun address|

sw $t7,12($sp)         #$t7 |from |

sw $a1,8($sp)         #$a1 |to |

sw $a2,4($sp)         #$a2 |numberof disc|

sw $t1,0($sp)         #$t1 |temp |<---$sp



bne $a2,1,recurse # if (n =/= 1) recurse



li $v0,4

la $a0,msg1

syscall

li $v0,1

move $a0,$t7

syscall

li $v0,4

la $a0,msg4

syscall

li $v0,1

move $a0,$a1

syscall

li $v0,4

la $a0,msg5

syscall



endcase: # exit code



addi $sp,$sp,20         #delete a stack with 5 boxes



lw $ra,16($sp)         #$ra |retun address|

lw $t7,12($sp)         #$t7 |from |

lw $a1,8($sp)         #$a1 |to |

lw $a2,4($sp)         #$a2 |numberof disc|

lw $t1,0($sp)         #$t1 |temp |<---$sp



jr $ra # return



recurse:



addi $a2,$a2,-1



move $a1,$t1



jal hanoi # call hanoi



li $v0,4        #--------------------(**)

la $a0,msg2

syscall

li $v0,1

move $a0,$a2

syscall

li $v0,4

la $a0,msg3

syscall

li $v0,1

move $a0,$t7

syscall

li $v0,4

la $a0,msg4

syscall

li $v0,1

move $a0,$a1

syscall

li $v0,4

la $a0,msg5

syscall



lw $ra,16($sp)         #$ra |retun address|

lw $t7,12($sp)         #$t7 |from |

lw $a1,8($sp)         #$a1 |to |

lw $a2,4($sp)         #$a2 |numberof disc|

lw $t1,0($sp)         #$t1 |temp |<---$sp

addi $sp,$sp,20         #delete a stack with 5 boxes



addi $a2,$a2,-1



move $t7,$t1





jal hanoi # call hanoi



j endcase # goto exit code (return)---------------------(***)

help understand code

shahab03 — Mon, 05 Oct 2009 18:39:59 GMT

I have following code which reverses any input provided using recursion. I am trying to understand what each line of code does. so I can have a better understanding.

so I have some questions, wondering if someone can help me understand whats happening in the code. I have commented each line where I am a question. any help will be greatly appreciated.

I don't understand anything under the label notend. please share any thoughts you might have .. thanks

.data

mess1: .asciiz "Enter string: "
mess2: .asciiz "Reversed string: "
input: .space 128
newln: .asciiz "\n\n"
output: .space 2
.text
.align 2

main:
li $v0, 4 #print string
la $a0, mess1
syscall

li $v0, 8 #read string
la $a0, input #pointing to 128(bytes) of space reserved by input
la $a1, 127 #read upto 127 characters
syscall

li $v0, 4 #print string
la $a0, newln
syscall

li $v0, 4 #print string
la $a0, mess2
syscall

la $a0, output #output will be single character
addi $a0,$a0,1 -- whats in a0 at this point and why are we adding??
add $a1,$zero,$zero -- whats happening here?? why we setting $a1 to zero?

sb $a1, 0($a0) #set null at the end of output -- dont undertand ??
la $a0, input #prepare for proc call ??
li $a1, 0 # $a1 contains index ??
jal rev

rev:
add $t1,$a1,$a0

lbu $t0, 0($t1) #load cur chacter -- what does 0($t1) mean?
bne $t0, 0xa, notend # what is 0xa ??

jr $ra # which line is this jumping to ?? line 33 sb $a1, 0($a0) ??

notend:

addi $sp, $sp, -8 # whats happening here? why substract 8 from stack pointer?

sw $ra, 4($sp) # save return address -- why?

sw $t0, 0($sp) # save cur char

addi $a1, $a1, 1 # increase index

jal rev

la $a0, output # stuff the cur char into the 'output'

lw $a1, 0($sp)

sb $a1, 0($a0)

li $v0, 4 # print current char

la $a0, output

syscall
lw $ra, 4($sp) # restore caller state

addi $sp, $sp, 8 # free up the stack space used

jr $ra

Tables and sorting

tyczj — Mon, 05 Oct 2009 17:25:46 GMT

so i need to input 10 ints into a table then use insertion sort but i dont know how to step through the table like you would and array in c++ since i have found out i cant go tbl[length-1] for example because i get an error saying "cant add relative quantities".

in short how do i go from number to number in the table to step through it

Creating a floppy image

makaveli0129 — Mon, 05 Oct 2009 09:46:03 GMT

Ok so i have this floppy image that just prints a message now the problem that i am running into is that i have to jump to memory location 0x1000 and then jump back to the original place but have no clue how to do it this is what mbr looks like on the floppy: (i am using nasm)

org 0x7c00

xor ax,ax

mov es,ax

mov ah,0

mov al,3

int 10h



mov ah,13h

mov al,1

mov bh,0

mov bl,0ah

mov cx,mlen

mov dh,0

mov dl,0

mov bp, msg

int 10h



;jump to 0x1000 here

;print '$' here



mov dh,1

msg db "This is a message"

mlen equ $-msg

times 512-($-$$)-2 db 0

dw 0AA55h

Problem with MIPS

ripineros — Mon, 05 Oct 2009 06:14:53 GMT

Hi everyone,

We are learning MIPS at one of my classes and in the book there are some exercises, but I'm having problem with one of them. It is a buggy code and we are supposed to find the cause of an exception.

# This code is supposed to calculate A*X + B

# for N values of X



# A and B are in $a1 and $a2, respectively

# the X values and N are in the data segment

#

        

        .text

main:        

        li $a1, 3

        li $a2, 4

        la $t0, N

        la $t1, X

        lw $t2, 0($t0)



        xor $t3, $t3, $t3

        xor $t4, $t4, $t4



loop:        addi $t2, $t2, -1

        blez $t2, exit

        lw $t3, 0($t1)

        mul $t3, $a1, $t3

        add $t3, $a2, $t3

 

        li $v0, 4

        la $a0, outputMsg

        syscall

        li $v0, 1

        add $a0, $0, $t3

        syscall

        li $v0, 4

        la $a0, newln

        syscall



        add $t1, $t1, 1

        j loop



exit:        

        jr $ra



        .data



N:        .word 4

X:        .word 12, 14, 16, 18

         

outputMsg:

        .asciiz        "\n Result = "

newln:

        .asciiz "\n\n"

I tried single stepping the code, and the exception seems to happen always when executing this instruction after the first loop.

lw $t3, 0($t1)

Whats wrong with $t1?

Thanks for the help.

Program help required

bilalqasim — Mon, 05 Oct 2009 04:46:08 GMT

Please can anyone help me in the making of this program :

We have to Enter a string of capital letters like

WJKTYNHJHIJKLMNOPASC

THE LONGEST CONSECUTIVELY INCREASING STRING IS

HIJKLMNOP

.
.
.

Help me Please , Thanks !!!

question?

jingo1126 — Sun, 04 Oct 2009 11:12:25 GMT

guys?is it even possible that once i created a file already with my program then execute that program again wont create a file anymore?.. like q.txt then its created..but when i execute my program again and type a string to create a file like w.txt..it doesnt create the file..theres a screenshot of what ive done with my sample codes..it kept bugging making me crazy to figure out whats wrong..
heres my code if there is something wrong please help me:

.model small

.stack 100h

.data

buff db 10,?,10 dup(' ')

handle dw ?

.code

start:

mov ax,seg buff

mov ds,ax

mov dx,offset buff

mov ah,0ah                    ; string input

int 21h

push ax                          ; string placed to stack

je print

                print:

                xor bx, bx

                mov bl, buff[1]

                mov buff[bx+2], '$'

                mov dx, offset buff + 2

                mov ah, 9

                int 21h

mov ah,3ch                    ; create file attribute

mov cx,0                        ; normal attribute

pop bx                           ; get data from stack

mov ah,3ch                    ; create file with file name from stack

int 21h

mov handle,ax

finish:

mov ah,4ch                    ; terminate program

int 21h

end start

Attached Images

Capture.JPG (71.9 KB)

working or not?need help

jingo1126 — Sat, 03 Oct 2009 22:53:50 GMT

first things first..i still have my exe file that the program works..
i was just putting labels to my program then the next thing happens it does not work anymore except for the terminate program..
if some of you guys have time please recomplie it and test it in your tasm.. these are all my codes.. please help me

;Jingo Rodriguez

;Project Partial File Manager Creates File, Directory and Deletes

;has better interface except no int on keyboard and and mouse int whch are 16 and 33

;ints use are 10h and 21h

;one flaw in program is you have to execute it again when an error is made..like wrong spelling and wrong input of data at the beginning..just like the one your experiencing in wrong or false use of program

.model small

.stack 100h

.data

buffer db 10,?, 10 dup(' ')

handle dw ?

.code

start:

call clear                                        ;call clear procedure call is placed when theres a certain procedure to be called

started:

mov ah,1                                        ;service 1h is character input with echo

int 21h

push ax                                                ;stores value at ax for further use



pop bx                      ;gets the value from stack for usage now the stack is emptied...



cmp al,"F"                  ;checks if the desired character has been input if not proceeds to then next line till it gets the desired input then jumps to a procedure or label

je file                                                ;use je instead of call why?because call may run the procedures leaving other procedures useless.



cmp al,"o"                                        ;" "

je folder



cmp al,"D"                                        ;" "

je deletes



cmp al,"x"                                         ;jumps to finishs                        

je finishs



file proc                                        

                mov dl,0ah          ; linefeed

                mov ah,2            ; service 2h character output

                int 21h  

                mov ax,seg buffer           

                mov ds,ax                   

                mov dx, offset buffer

                mov ah, 0ah                        ; string input is what ive learned most in this program then storing it at stack

                int 21h

                push ax                                ; string stored at stack                

                jmp print

                print:                                ; print what is been inputed maybe need or may not be needed at all.

                xor bx, bx

                mov bl, buffer[1]

                mov buffer[bx+2], '$'

                mov dx, offset buffer + 2

                mov ah, 9           ; service 9h is for string output

                int 21h

crtfle:

mov ah,3ch

mov cx,0                  ;  is best use to check if the 3ch service is working..change maybe done if neccessary for the program

 ;mov cx, 0       ;  normal - no attributes. 

 ;mov cx, 1       ;  read-only. 

 ;mov cx, 2       ;  hidden. 

 ;mov cx, 4       ;  system 

 ;mov cx, 7       ;  hidden, system and read-only! 

 ;mov cx, 16      ;  archive 

pop bx                                                

mov ah,3ch

int 21h

mov handle, ax              

jmp finish

file endp                                        

folder proc

                mov dl,0ah                          ;linefeed

                mov ah,2                    ;service 2h character output

                int 21h                    

                mov ax,seg buffer  

                mov ds,ax

                mov dx, offset buffer  

                mov ah, 0ah         ;service use are 0ah of int 21h: is use to input string

                int 21h

                push ax             ;after input string will be push to register ax

                jmp print1

                print1:             ;code is to print the string after input

                xor bx, bx

                mov bl, buffer[1]

                mov buffer[bx+2], '$'

                mov dx, offset buffer + 2

                mov ah, 9

                int 21h

                pop bx

                mov ah,39h           ;create folder at current directory

                int 21h

                jmp finish

                finishs:             ;this jumps are made due to the fact they can't jump a long jump that exceeds at 128 bytes..

                jmp finish           ;short jmps are but will make it a bit slower due to some sources i found on the net

                deletes:                    ;this jump will jump from one another till it goes to the desired procedure or label.

                jmp delete

folder endp

delete proc

                mov dl,0ah                         ;linefeed

                mov ah,2             ;service 2h character output

                int 21h

                mov ax,seg buffer  

                mov ds,ax

                lea dx, buffer  

                mov ah, 0ah          ;service use are 0ah of int 21h: is use to input string

                int 21h

                push ax              ;after input string will be push to register ax

                jmp print2

                print2:                      ;code is to print the string after input

                xor bx, bx

                mov bl, buffer[1] 

                mov buffer[bx+2], '$'

                mov dx, offset buffer + 2

                mov ah, 9

                int 21h

                pop bx

                mov ah,41h           ;delete file at current directory

                int 21h

        jmp finish

delete endp

Clear proc ;only a design for better viewing..UI(User interface)

;learning putting color at specific places at screen is nice, screen of dos has only 80 pixels width

;ok i can understand almost everyline except for the two line below 

;which means i guess they might come in handy for some reasons maybe trying to remove something place at stack so that when procedure is being called it has a clean data to throw in the process.

; set es (just in case): 

push    cs

pop     es



mov     bh, 0    ; page. 

lea     bp, tmsg  ; offset. 

mov     bl, 01fh ; default attribute. 

mov     cx, 80   ; char number. 

mov     dl, 0    ; col. 

mov     dh, 0    ; row. 

mov     ah, 13h  ; function. 

mov     al, 1    ; sub-function. 

int     10h



mov     bh, 0    ; page. 

lea     bp, msg  ; offset. 

mov     bl, 01fh ; default attribute. 

mov     cx, 80   ; char number. 

mov     dl, 0    ; col. 

mov     dh, 1    ; row. 

mov     ah, 13h  ; function. 

mov     al, 1    ; sub-function. 

int     10h





mov     bh, 0    ; page. 

lea     bp, cmsg ; offset of string with attributes. 

mov     bl, 01fh ; default attribute (not used when al=3). 

mov     cx, 80   ; char number. 

mov     dl, 0    ; col. 

mov     dh, 2    ; row. 

mov     ah, 13h  ; function. 

mov     al, 1    ; sub-function. 

int     10h





mov     bh, 0    ; page. 

lea     bp, opt1 ; offset of string with attributes. 

mov     bl, 01fh ; default attribute (not used when al=3). 

mov     cx, 8   ; char number. 

mov     dl, 22    ; col. 

mov     dh, 3    ; row. 

mov     ah, 13h  ; function. 

mov     al, 3    ; sub-function. 

int     10h



mov     bh, 0    ; page. 

lea     bp, opt2 ; offset of string with attributes. 

mov     bl, 01fh ; default attribute (not used when al=3). 

mov     cx, 9   ; char number. 

mov     dl, 30   ; col. 

mov     dh, 3    ; row. 

mov     ah, 13h  ; function. 

mov     al, 3    ; sub-function. 

int     10h

mov     bh, 0    ; page. 

lea     bp, opt3 ; offset of string with attributes. 

mov     bl, 01fh ; default attribute (not used when al=3). 

mov     cx, 8   ; char number. 

mov     dl, 37    ; col. 

mov     dh, 3    ; row. 

mov     ah, 13h  ; function. 

mov     al, 3    ; sub-function. 

int     10h



mov     bh, 0    ; page. 

lea     bp, opt4 ; offset of string with attributes. 

mov     bl, 01fh ; default attribute (not used when al=3). 

mov     cx, 8    ; char number. 

mov     dl, 45   ; col. 

mov     dh, 3    ; row. 

mov     ah, 13h  ; function. 

mov     al, 3    ; sub-function. 

int     10h



mov     bh, 0    ; page. 

lea     bp, left ; offset of string with attributes. 

mov     bl, 011h ; default attribute (not used when al=3). 

mov     cx, 22   ; char number. 

mov     dl, 0    ; col. 

mov     dh, 3    ; row. 

mov     ah, 13h  ; function. 

mov     al, 1    ; sub-function. 

int     10h



mov     bh, 0    ; page. 

lea     bp, right ; offset of string with attributes. 

mov     bl, 011h ; default attribute (not used when al=3). 

mov     cx, 27   ; char number. 

mov     dl, 53   ; col. 

mov     dh, 3    ; row. 

mov     ah, 13h  ; function. 

mov     al, 1    ; sub-function. 

int     10h



mov     bh, 0    ; page. 

lea     bp, inst  ; offset. 

mov     bl, 01fh ; default attribute. 

mov     cx, 80   ; char number. 

mov     dl, 0    ; col. 

mov     dh, 4    ; row. 

mov     ah, 13h  ; function. 

mov     al, 1    ; sub-function. 

int     10h

ret  ; return control to the operating system

tmsg db '                                                                                 ' ;design top most of the screen



 msg db 'Rodriguez File Manager Version 0.1.1                                             ' ;design second to the top followed by the rest

        

cmsg db  '__________________________C h o o s e  a n  O p t i o n__________________________'

     

opt1  db ' ',01fh, 'F',003h, 'i',030h, 'l',030h, 'e',030h, ' ',01fh, ' ',01fh ,' ',01fh

opt2  db 'F',030h, 'o',003h, 'l',030h, 'd',030h, 'e',030h, 'r',030h ,' ',01fh ,' ',01fh

opt3  db ' ',01fh, ' ',01fh, 'D',003h, 'e',030h, 'l',030h, 'e',030h, 't',030h, 'e',030h 

opt4  db ' ',01fh, ' ',01fh, ' ',01fh, 'E',030h, 'x',003h, 'i',030h, 't',030h, ' ',01fh

left  db '                   '

right db '                         '

inst  db 'Preess the letter if the letter is capital, press shift and the letter           '

Clear endp

finish:

mov ah,4ch ; terminate program

int 21h

end start

anyone know how to encrypt?

jingo1126 — Fri, 02 Oct 2009 06:27:55 GMT

guys wondering how come the code doesnt work..
im using tasm as u might know..im checking on things work and want to see what its output be..thanks hope you can help me with this one. i tried converting it tasm but say argument needs to be override.

.model small

.stack 100h

.data

; string has '$' in the end: 

string1 db 'hello world!', 0Dh,0Ah, '$'

;                       'abcdefghijklmnopqrstvuwxyz' 

table1 db 97 dup (' '), 'klmnxyzabcopqrstvuwdefghij'

table2 db 97 dup (' '), 'hijtuvwxyzabcdklmnoprqsefg'

.code

start:

; encrypt: 

lea bx, table1

lea si, string1

call parse

; show result: 

lea dx, string1

; output of a string at ds:dx 

mov ah, 9

int 21h

; decrypt: 

lea bx, table2

lea si, string1

call parse

; show result: 

lea dx, string1

; output of a string at ds:dx 

mov ah, 09

int 21h

ret   ; exit to operating system. 



; subroutine to encrypt/decrypt 

; parameters:  

;             si - address of string to encrypt 

;             bx - table to use. 

parse proc near



next_char:

        cmp [si], '$'

        je end_of_string

        mov al, [si]

        cmp al, 'a'

        jb  skip

        cmp al, 'z'

        ja  skip

        ; xlat algorithm: al = ds:[bx + unsigned al]  

        xlatb     ; encrypt using table2.   

        mov [si], al

skip:

        inc si

        jmp next_char

end_of_string:

ret

parse endp

end start

printing a digit (as a character)

andreib — Thu, 01 Oct 2009 19:51:55 GMT

Hello!
I'm kind of new in assembly programming and I've been struggling with the following problem today:
How to print an integer, using BIOS interrupt 10h, with its 0Eh in AH register, as its function to print a character to the screen...

For example, how to print the last digit of 2005.
here's what I've tried, using division by 10:

mov ax, 2005

mov cx, 10

div cx     ; now, AH should contain the remainder (5) and AL = 200

mov al, ah ; because AL should contain the character to print

add al, 48 ; because ASCII code for 5 is 53 (decimal).

mov ah, 0Eh; the function for the 10h interrupt

int 10h

I can't understand why 0 is printed to the screen instead of 5. If i change the line

add al, 48

to

add al, 49

it prints out 1.
... am I missing something? is it because I use decimal format for integers instead of the hexadecimal one?

Thank you in advance!

Unsure of AT&T syntax and segfault

chrisname — Thu, 01 Oct 2009 19:47:49 GMT

I'm actually writing this code as inline ASM but I feel it is more relevant to ASM than it is to C...

Basically all I want to do for now is print the alphabet. However I'm having some trouble getting used to AT&T syntax (I would rather learn it than keep just using Intel syntax) and I'm also unsure as to how to solve this segfault I'm getting:

#include 



int main() {

    // code

    {

        asm volatile("movb %ah, 0x4C\n" /* BIOS interrupt call for output */

            "movb $'A', %AL\n"          /* Place "A" in AL                */

            "jmp start\n"

            "start:\n\t"

                "add $1, %AL\n\t"   /* Add 1 to 'A' to move on to 'B', etc. */

                "print:\n\t\t"

                    "int $0x21\n\t" /* Print contents of AL */

              #if 0

                "hlt\n"       /* Halt the machine (commented out because I think I need ring 0 access to use hlt) */

              : " %ax", "%bx" /* List of registers used (commented out because GCC complained of the comma)       */

              #endif

            );

    }

    // code

    return 0;

}

I have tried without the "volatile" keyword. I did nothing, as I expected...

Thank you :)

String to integer

ov3rcl0ck — Thu, 01 Oct 2009 02:42:07 GMT

So when I take an argument off the stack it seems that is in the form of a string, is there a way i can convert it to a integer so i can "cmp" it with another argument which i also wish to convert to an int? I am a newbie to ASM but i do realize this isn't going to be as easy as it is in higher level languages. Any help?

read and evaluate string in spim/mips

shahab03 — Wed, 30 Sep 2009 17:49:40 GMT

I believe I can read string using

li $v0, 8
syscall

how can I take what was read and check if it was a specific value.

e.g.
if input was Y then jump to loop label. if it was N then jump to exit label.

thanks

Font color code

geno93n0 — Wed, 30 Sep 2009 13:14:38 GMT

Is there a code to set the font color for all outputted text in a86? I have to combine the following two programs, using the background from SIEGE and the content of FTCONV. My problem is that I'm working on a limited timeframe and I'd really like to know if there's something that can speed up my job, which is to make all the text in FTCONV white when I combine it with SIEGE.

Any ideas? Files attached below for perusal.

Attached Files

	SIEGE.txt (19.0 KB)
	ftconv.txt (13.7 KB)

Code Snippet Passive TSR Hello World

NotNull — Tue, 29 Sep 2009 19:21:10 GMT

Assembles under NASM 16-bit,
invoke INT 0x65 to display message.

OS dev: Procedure that ends a task.

Alex_ — Tue, 29 Sep 2009 09:49:37 GMT

Hello.
I have an assignment to make a procedure that will end a task. The manual from the university isn't quite clear about how this works. So mainly i have this problem: How does a procedure end a working task?
1) Does the bios call the procedure ? or,
2) Will my current operating system call my procedure? or,
3) I must make myself an os that boots, launches a task, and then after an amount of time calls the procedure to end the task and bail out.
I have no idea!
The manual does give some sort of code(which should be changed by me to work properly) that shows how it should look like. But once i don't understand the way it works, how can i know what to change?
The code with comments (translated by me) from the manual:

PROC        newint20h                ; int 20h

        call        _di



;keeping back any interruptions so that the task may not be launched when the processor is taken away (me: what does "taken away" mean????)

         pop        ax

        pop        ax        ;Extracting from stiva (me: I don't know the true translation of stiva. But one thing for sure: it's a FILO type memory acces)

        pop        ax



;extracting the leftovers that are left from the execution of int 20h (we don't need the CS,IP indicators any more) (me: What int 20h? Are they talking about THIS procedure or the system's one.)

        push         cs

        pop        ds

        lea        di,[cs:firsttask]        ; DI = Offset of TASK

        mov        al,[cs:numtask]        ; AL = Task's ID

        xor        ah,ah                ; AX = Task's ID

        push         dx

        mov        dx,515

        mul        dx        ; AX = the starting index of firsttask

        pop        dx

        add        di,ax        ; DI = putting the top of the stiva in di



;Putting a 255 value into task's ID , marking it as non-loaded.

        mov        [Byte Ptr cs:di],255

        mov        dx,[cs:di+3]

        call        FreeMemory

        call        _ei

        jmp        int08new

; Freeing the memory, and beginning the execution of int 8;

        ENDP        newint20