Write data to a char using Assembly

Question

I created a program in C to generate a char* processedData. I send to my assembly program and put it in a register:

mov     edx, [ebp+12]
mov     edi, edx

How can i write a char into it. I know i need to Write a char and inc edi... that in a loop. But how can i write a char, i already have the value into another register. But if i do mov edx, 49; char code i'll lose the pointer. I want to do something like

for(p=malloc(100*sizeof(char*)); p!=NULL;p++){
    *p=//my char code
}

Assembly for linux (DEBIAN) x86

Answer 1

edx is the address of the destination of the char. That is, edx is a pointer to the location you want to write to. Therefore, do this:

mov byte ptr ds:[edx], 49

This might even work:

mov byte ptr ds:[edx], '1'

You say your character is already in some register. I assume it is an 8 bit register ( ah/al/bh/bl/ch/cl/dh/dl ), in which case you can just do:

mov [edx], ah

The assembler can infer the size of the data [edx] is pointing to in this case.

---

I'm on Windows right now so this code is for VC++. It demonstrates copying a string:

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

char src[] = "hello";
char dest[8];

int main( void )
{
   __asm
   {
      xor ecx, ecx

      mov eax, offset src
      mov edx, offset dest

        loop_dest:

      mov bh, byte ptr ds:[eax+ecx]
      mov [edx+ecx], bh

      inc ecx
      cmp ecx, size src
      jnz loop_dest
   }

   printf("%s\n", dest);
   return EXIT_SUCCESS;
}

If your assembler uses AT&T syntax, you'll need to do some minor translation, but hopefully this points you in the right direction.