Accessing C pointers from assembly

Question

I am currently learning how to intermix C code with assembly. In the .c file I have the following code:

extern int replace(char *c); int main(){ static char * string2 = "aaab\n"; static char * string3; string3 = (char *) replace(string2); *string3 = 'A'; printf("%s",string3); }

The assembly function replace has the following code:

section .data string db "Test",10,0 section .text global replace replace: enter 0,0 mov eax, string leave ret

This works very well and shows that the C program has access to the .asm file memory. However when I try to write to the pointer string2, say by adding the following code to the function replace,

mov edx, [ebp + 8] mov byte [edx], 10

I obtain a segmentation fault. (From my understanding the latter is passing the address of the pointer to edx, thus edx = string2, and then assigning *string2 = 10 via *edx = 10). Could you please explain what I am doing wrong?

EDIT: I figured out that when I do a string3 = (char *) malloc(10); then I can indeed do in assembly

mov edx, [ebp + 8] move byte [edx], 10

and obtain correct results. I would still be grateful for an explanation of why is this so. Is this due to the heap being publicly accessible? Is it perhaps because before char *string = "Hi" was pointing to a read-only part of the memory where strings are stored? I am learning this on my own so I would be grateful for clarifications.

David C. Rankin · Accepted Answer · 2015-07-09 04:50:43Z

Your problem in your C program is due to creating a string literal that cannot be modified with:

static char *string2 = "aaab\n";

You assign pointer string3 the address of string2 via your replace call:

string3 = (char *) replace(string2);

You then attempt to modify read-only memory (string2) by changing the first character from a to A with:

*string3 = 'A';

Which results in the segfault. To fix the segfault, just change the string2 declaration to:

static char string2[] = "aaab\n";

So that it is simply a character array and not a string literal.

When you create a literal with char *label = "stuff"; you create a literal in read only memory. You can easily see this for yourself by dumping the assembly of the C program with gcc -S -masm=intel -o file.asm file.c. If you look, you will see that "aaab\n" is created in the .rodata (read-only) data section:

 .section .rodata .string "aaab\n" .data .align 8

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