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I am learning binary exploitation from "Hacking: The Art of Exploitation". In theory the address of an element of stack should be higher than the address of an element on heap. But in this program I am getting the address of stack_var is lower than that of heap_var. I am using Ubuntu 18.04.2 LTS with AMD Ryzen 5 2500u processor.

#include <stdio.h>

int global_var;
int global_initialized_var = 5;

void function(){
    int stack_var;

    printf("the function's stack_var is at address 0x%08x\n", &stack_var);
}

int main(){
    int stack_var;
    static int static_initialized_var = 5;
    static int static_var;
    int *heap_var_ptr;

    heap_var_ptr = (int *) malloc(4);

    //These variables are in data segment
    printf("global_initialized_var is at address 0x%08x\n", &global_initialized_var);
    printf("static_initialized_var is at address 0x%08x\n", &static_initialized_var);

    //These variables are in bss segment
    printf("static_var is at address 0x%08x\n", &static_var);
    printf("global_var is at address 0x%08x\n", &global_var);

    //This variable is in heap segment
    printf("heap_var is at address 0x%08x\n", heap_var_ptr);

    //These variables are in stack segment
    printf("stack_var is at address 0x%08x\n", &stack_var);
    function();
}

Output:

global_initialized_var is at address 0x91d2f010
static_initialized_var is at address 0x91d2f014
static_var is at address 0x91d2f01c
global_var is at address 0x91d2f020
heap_var is at address 0x9292b260
stack_var is at address 0x242e2bec
the function's stack_var is at address 0x242e2bc4
  • I guess this is to do with ASLR. What's your compiler and what are compiler options? – domen Jul 22 '19 at 8:33
  • @domen It is GNU compiler collection(gcc) and compiled program with gcc memory_segment.c – coder Jul 22 '19 at 8:36
  • 1
    Would you address the compiler warnings first, and then check if your premise still holds? ;) – domen Jul 22 '19 at 8:41
  • @domen Now I am getting stack address higher than heap address. Used %p to print address instead of 0x%08x. – coder Jul 22 '19 at 8:55
2

As you have spotted, with %p it works as expected.

The explanation at the source of your situation is that %x reads integers (whose size is 4 bytes) out from pointers (whose size is 8 bytes on your architecture). So, you are simply asking printf to read the first 4 bytes of the pointer, neglecting the rest 4. You may change your format to "0x%016lx" to format pointers with leading zeros.

In those cases where the pointer is small enough to fit in 4 bytes, you see them (haphazardly) correctly because the least significant bytes are stored as first due to the fact that your architecture is little-endian.

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