Using the last two digits of input only.

[DeGeeGin]

Hi forum, I am just wondering how I could just use the last two digits of a four digit year the user inputs. For example, I want to drop the [19] from [1960] and just use the [60] which should be 3C in Hex so the registers would read it.

Here is my Data Segment.

NUMPAR		LABEL	    BYTE
MAXLEN		DB	    5		;maximum input length
ACTLEN		DB	    ?
NUMFLD		DB	    8 DUP(' ')
BINFLD		DW	    0
MULT10		DW	    1

And here is my code to accept ASCII characters from the user and convert them to digits. The user should input a four-digit year.

;To accept number.
		MOV	    CX,0
		MOV	    CL,MAXLEN
		LEA	    SI,NUMFLD
CLRIN:
		MOV	    BYTE PTR[SI],020H
		INC	    SI
		LOOP	    CLRIN
;
		MOV	    MULT10,1
		MOV	    BINFLD,0
		MOV	    AH,0AH
		LEA	    DX,NUMPAR
		INT	    21H
		MOV	    CX,10
		LEA	    SI,NUMFLD-1
		MOV	    BL,ACTLEN
		MOV	    BH,0
INPUTNXT:
		MOV	    AL,[SI+BX]
		AND	    AX,000FH
		MUL	    MULT10
		ADD	    BINFLD,AX
		MOV	    AX,MULT10
		MUL	    CX
		MOV	    MULT10,AX
		DEC	    BX
		JNZ	    INPUTNXT
;End input.

[gusano79]

To paraphrase, you want your INPUTNXT loop to only consider the last two characters entered, right?

You already have the information you need to find where those two characters start within NUMFLD : The user entered ACTLEN out of MAXLEN possible characters, so there are ( MAXLEN - ACTLEN ) extra characters at the start that you don't care about. Rather than start converting characters at NUMFLD , use that value as your starting offset.

[thines01]

That really depends on if you want the hex value or a string representation:
This one gets the text value that can be converted to a byte (if necessary);
Pardon the weird labels.

; Compiled with A86 (http://eji.com/a86/)
; This program shows how to get user input with echo.

ORG 100H
	mov ah, 09h
	mov dx, askName;
	int 21h                      ; Ask for year
	
	mov cx, 0ah
	mov si, userInput;            ; Place to store the user name
getName:
	mov ah, 01h
	int 21h                      ; Get input with echo
	
	cmp al, 0dh                  ; 
	je printQuit                 ; Quit if Carriage-return
	
	mov [si], al                 ; Put entered char in user name
	
	cmp byte ptr [finalStop], 24h; is 4th dollar-sign gone?
	jne printQuit                ; if so, buffer is full?, quit
	
	inc si			     ; move the position
	loop getName;		     ; Keep going (looping)
printQuit:
	mov ah, 09h
	mov dx, youEntered;
	int 21h			     ; Print the finished input
	mov dx, lastDigsLabel;
	int 21h
	mov dx, userInput+2;
	int 21h
	
quit:
	int 20h                      ; *** DROP TO DOS ***
	
askName:	db 'Enter a 4-digit year and press enter: $'
youEntered:	db 'You entered', 0dh, 0ah
userInput:	db '$$$'
finalStop:	db '$$'
lastDigsLabel:	db 0dh, 0ah, 'Last two digits are: $'

[DeGeeGin]

I figured it out, but thanks though. All I needed to do was divide the date by 100 and take the remainder... So simple... >.