Question

I have a very basic question.
Lets take this snippet:

#include <stdio.h>

void foo(void) {
    char *s = "StackOverflow";
    printf("%s\n", s);
}

int main(void) {
    foo();
}

In the process execution stack, main gets loaded on to the stack, then foo() gets called.
Now, where is the memory for “StackOverflow” allocated?
Similarly where is the memroy for “%s\n” allocated when printf is called?

---

Consider the following code:

Now the other question I have is, considering the below code:

#include <stdio.h>

int x;
int abc = 100;

void foo(void) {
    char *s = "stackoverflow";
    printf("%s\n", s);
}

int main(void) {
    foo();
}

So, if I do objdump -s -j .bss a.out , I should see uninitialized segment and if I do objdump -s -j .data a.out , I should see initialized segment (abc=100) rt? Is there anything wrong with this assumption?

I get the following outputs though:

test > objdump -s -j .bss a.out
a.out: file format elf32-i386

test > objdump -s -j .data a.out

a.out: file format elf32-i386

Contents of section .data:
804954c 00000000 3c960408 00000000 64000000 ….<…….d…

What am I missing here?

thanks everyone again

Answer 1

"StackOverflow" and "%s\n" string literals are put in .rodata (read only data ) section in most systems.

On UNIX, you can dump .rodata section using the objdump command:

$ gcc tst.c 
$ objdump -s -j .rodata a.out

As added by @FatalError in the comments, "%s\n" is not visible with objdump in the example as gcc optimizes a call to printf("%s\n",str) by replacing it by a call to puts(str).

To see the "%s\n" string literal in the objdump output, you can compile your program with gcc -fno-builtin.