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Hello Gyes!
I am very new in Using emu8086 and i need a code about this Program Problem:

I have to convert a Hexa input to a Binary Output using emu8086.
1. Hexa number will be placed in AL
2.And output Binary will come in AX+ DX, Memory address will be 0507 to 0500

I tried to do but last of all i mixed up everything...


Plz Guys help me to do this problem. Thank you ..

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Dani AI

Generated

A few focused notes on the attempts by and a clearer, reliable approach for 8086 emu8086.

The main problem in the posted attempts is mixing division semantics and reconstruction math. On 8086 the DIV instruction size matters: an 8-bit divisor divides AX (quotient → AL, remainder → AH); a 16-bit divisor divides DX:AX (quotient → AX, remainder → DX). If AX/AH or DX are not set exactly as required, remainders and quotients will be garbage — which explains the wrong bits seen. Clearing the high byte (AH) or using the correct divisor width is mandatory when trying repeated DIV-by-2 loops.

A simpler, safer method is bit extraction rather than arithmetic division. Treat the byte as eight independent bits and write each bit as a byte ('0' or '1') into a small output buffer. Two common, easy-to-reason-about ways work well on 8086: test with masks (80h, 40h, 20h, ... 01h) from MSB to LSB, or shift the working byte and read the carry/low bit each iteration. Both avoid multiword division, avoid rebuilding values with decimal/hex place multiplications, and produce predictable output order for display.

Practical cautions and test tips: choose whether the buffer holds ASCII '0'/'1' (0x30/0x31) or numeric 0/1; initialize segment registers or use labels so writes go to the expected memory; single-step in the emulator and watch AX/AH/DX when trialing DIV; inspect the buffer in memory view to confirm order. Using a small, clearly indexed loop for eight iterations will be far more robust than repeated DIV and multiply steps.

Member Avatar for mahafuz_eee

Please kindly help me.

Member Avatar for mahafuz_eee

MOV AL, 0AH
MOV BX, 2H

MOV DX, 0000H
DIV BX

MOV [0500], DL
MOV DX,0000H
DIV BX

MOV [0501], DL
MOV DX, 0000H

DIV BX
MOV [0502], DL

MOV DX, 0000H
DIV BX

MOV DX, 0000H
MOV [0503], DL
DIV BX

MOV AX, [0500]
MOV BX, 1H
MUL BX

ADD [0504], AX
MOV AX, [0501]
MOV BX, 10H
MUL BX

ADD [0504], AX
MOV AX, [0502]
MOV BX, 100H
MUL BX

ADD [0504], AX
MOV AX, [0503]
MOV BX, 1000H
MUL BX

ADD [0504], AX
MOV AX, [0504]

RET


This is the Code. Buit in Th AX Register i Found 0110 But it Should be 1010.
PLEASE BROTHER ANY ONE HELP ME

Member Avatar for mahafuz_eee

; to convert a hexa number to a binary number

mov al,0ch
mov bl,2h

mov dx, 0000h
div bx
mov [0500],dl


mov dx, 0000h
div bx
mov [0501],dl

mov dx, 0000h
div bx
mov [0502],dl


mov dx, 0000h
div bx
mov [0503],dl


mov al, [0503]
mov bx, 1000h
mul bx
add [0508],ax


mov al, [0502]
mov bx, 100h
mul bx
add [0508],ax


mov al, [0501]
mov bl, 10h
mul bl
add [0508],ax


mov al, [0500]
mov bl, 1h
mul bl
add [0508],ax


mov ax,[0508]

ret

this is the solution. u all are chitter and dont even try to help in this simple program.

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