Question

I am programming in Tasm and want to input a 32 bit number.

I know that i have to input it digit by digit (i hope there is no one call function for number input)

This is my code

    .        .486
    .model small
    .code
    start:

    mov ebx, 0

    ; enter again and again untill enter is hit
    again:
    mov ah,01h
    int 21h
    cmp al, 13
    jz next
    mov dl, al
    mov eax, ebx
    mov ebx, 10
    mul ebx
    mov ebx, eax
    mov eax, 0
    mov al, dl
    add ebx, eax
    jmp again

    ; now find the digits back

    next:
    ; just testing to see if i got number
    mov edx, 13
    mov ah, 02h
    int 21h

    mov edx, 10
    mov ah,02h
    int 21h

    mov edx, ebx
    mov ah, 02h
    int 21h

    mov eax, ebx

    mov ebx, eax

    xor edx, edx
    xor cl, cl

    ; find digits and push into stack from last to first so when i pop i get digits back
    finddigit:
    xor edx,edx
    mov ch , 10
    div ch
    push dx ;taking risk dx dl
    inc cl
    cmp eax, 0
    jz print
    jmp finddigit


    ; stored into cl the number of digits

    print:
    cmp cl,0
    jz exit
    dec cl
    pop dx
    mov ah,02h
    int 21h
    jmp print


    exit:
    end start

i am stopping input at enter .

i am getting error NTVDM encountered a hard errror.

Thanks

This is my new modified code. It is working fine for some numbers like 2 and 123
but is failing for 333, 4444, 555; (i hope pushing and poping does not modify any registers other than specified):

.486
.model small
.code
start:

mov ebx, 0

; enter again and again untill enter is hit
again:
mov ah,01h
int 21h
cmp al, 13
jz next
mov cl, al
sub cl, 30h
mov eax, ebx
mov ebx, 10
mul ebx
mov ebx, eax
mov eax, 0
mov al, cl
add ebx, eax
jmp again

; now find the digits back

next:
; just testing to see if i got number
mov edx, 13
mov ah, 02h
int 21h

mov edx, 10
mov ah,02h
int 21h


mov eax, ebx

mov ebx, eax

xor ecx, ecx

mov ebx, ebp
; find digits and push into stack from last to first so when i pop i get digits back
finddigit:
xor edx,edx
mov ebp , 10
div ebp
push edx
inc cl
cmp eax, 0
jz print
jmp finddigit

; stored into cl the number of digits

print:
cmp cl,0
jz exit
dec cl
xor edx,edx
pop edx
add dl,30h
mov ah,02h
int 21h
jmp print


exit:
mov ah,4ch
int 21h               
end start

I am running this is MS-DOS CMD.exe window A pop up Error comes:

Answer 1

Assuming this is for DOS environment (due to int 21h):

There are some bugs in your code.

1. Read character function returns its output in al, that’s ok. But then you immediately destroy the value of read character with the following sequence:

    mov dl, al    ; the character read now in dl
    mov eax, ebx  ; eax is now 0 in the first loop
    mov ebx, 10   ; ebx is now 10
    mul ebx       ; the result of mul ebx is now in edx:eax,
                  ; the character read (in dl) is lost.

So you can’t store the character in dl if you are going to do mul ebx, because mul reg32 outputs the result in edx:eax. You could store it eg. in cl or ch instead.

2. Other bug which I notice is that you are attempting to multiply ASCII values by 10 (in the previous code). You should first subtract the value of ‘0’ of every read character before multiplication, that would be sub al, 30h or sub al, '0'.

3. Third bug is in the following sequence:

   xor edx,edx
   mov ch , 10
   div ch
   push dx ;taking risk dx dl
   inc cl
   cmp eax, 0
   jz print
   jmp finddigit

Edit: Here you are dividing ax by ch, which obviously does not work right for 32-bit division. As it seems that you want to have your dividend in eax, clear edx with xor edx, edx (as you do), and then divide edx:eax with a 32-bit register, eg. ebp, esi or edi (those seem to be unused in your code so far), and you’ll get the quotient in eax and the remainder in edx.