Question

The question consists of an assembly program that takes an input from a C program and divides it by a number, and returns the remainder to the C program to be printed as a string.

Here is my code for both:

#include <stdio.h>
#include <stdlib.h>

int main()
{
int i;
char *str;
str = malloc(1<<9);

printf("Enter a number: ");
scanf ("%d", &i);
printf("Number: %i\n", i);
str=int2string(i);
printf("Number as string is: %s\n", str);

return 0;

}

ASM =

%include "asm_io.inc"

segment .data

segment .bss

buffer resd 4   

segment .text
    global int2string
int2string:
    enter   0,0             ; setup routine
    pusha
    mov edx, 0

    mov eax, [ebp+8]        ; eax contains input value of int2string

    mov ebx, 10     ; sets ebx to value of 10
    div ebx         ; eax = eax / ebx

    call print_int      ; prints eax = quotient
    call print_nl       ; next line

    mov eax, edx        ; store edx (remainder) in eax

    call print_int      ; print remainder
    call print_nl       ; next line

    add eax, 48     ; convert result into ASCII character

    popa
    mov dword[buffer], eax  ; move ASCII character (if I replace eax with 48-57
                            ; it prints 0-9 correctly)
    mov eax, buffer     ; move buffer value to eax
    leave                     
    ret

My understanding is that the ASCII codes for numbers 0-9 is from the range 48-57, but if I don’t put the number explicitly in the buffer move, then the output is either garbage or segmentation fault.

Am I missing something here (eax with a value of 2 + 48, should be ASCII (50) = ‘2’)?

Answer 1

It’s likely that either print_int or print_nl may not preserve the value of eax (especially since a common calling convention is to return the result of a function in that register). If your second call to print_int prints the right value but the character is wrong, that’s a distinct possibility.

An easy test/fix for that is to change:

call print_int      ; print remainder
call print_nl       ; next line

into:

push eax
call print_int      ; print remainder
call print_nl       ; next line
pop eax

Other than that, it looks okay. All I can suggest is to single step through the code to see where it’s going wrong.