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```
org 100h
section .text
start:
mov bx, output ; put address of output into BX
mov ax, [num] ; put 16-bit value stored at num in AX
sub dx, dx ; let DX = 0
mov cx, 10000 ; Divide AX by 10000
div cx ; result in AX, remainder in DX
add al, 30h ; ASCII '1' = 49 (or 31h) so add 30h
mov [bx], al ; Store ASCII char in output
inc bx ; point to next char in output
mov ax, dx ; move remainder of last division into AX
sub dx, dx ; clear remainder
mov cx, 1000 ; Divide AX by 1000
div cx
add al, 30h
mov [bx], al
inc bx
mov ax, dx
sub dx, dx
mov cx, 100 ; Divide AX by 100
div cx
add al, 30h
mov [bx], al
inc bx
mov ax, dx
sub dx, dx
mov cx, 10 ; Divide AX by 10
div cx
add al, 30h
mov [bx], al
inc bx
mov ax, dx
sub dx, dx
mov cx, 1
div cx
add al, 30h
mov [bx], al
mov bx, output ; get address of first char in ouput
myloop:
mov dl, [bx] ; get char at address in BX
inc bx ; point BX to next char in message
cmp dl, 0 ; Is DL = null (that is, 0)?
je quit ; If yes, then quit
mov ah, 06 ; If no, then call int 21h service 06h
int 21h ; to print the character
jmp myloop ; Repeat for the next character
quit:
; DONE!
int 20h
section .data
num dw 12345
output db 0,0,0,0,0,0
```

Instead of dividing by 10000, then 1000, then 100,... the more normal way is to divide by 10 each time and output the bytes backwards - least significant first. This way, you can cope with numbers more flexibly. You then return a pointer to the most significant byte and only loop while you have digits left to do. I.e. your loop termination test (done after doing each digit) is AX equals zero. Doing the test after each digit means you will always output at least one digit - so a zero input causes a single digit zero output.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

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