I was following a tutorial that introduced stack overflows. Here is the c Code.

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

int main(int argc, char **argv)
  volatile int modified;
  char buffer[64];

  modified = 0;

  if(modified != 0) {
      printf("you have changed the 'modified' variable\n");
  } else {
      printf("Try again?\n");

and heres the disassembled main function

0x080483f4 <main+0>:    push   ebp
0x080483f5 <main+1>:    mov    ebp,esp
0x080483f7 <main+3>:    and    esp,0xfffffff0
0x080483fa <main+6>:    sub    esp,0x60
0x080483fd <main+9>:    mov    DWORD PTR [esp+0x5c],0x0
0x08048405 <main+17>:   lea    eax,[esp+0x1c]
0x08048409 <main+21>:   mov    DWORD PTR [esp],eax
0x0804840c <main+24>:   call   0x804830c <gets@plt>
0x08048411 <main+29>:   mov    eax,DWORD PTR [esp+0x5c]
0x08048415 <main+33>:   test   eax,eax
0x08048417 <main+35>:   je     0x8048427 <main+51>
0x08048419 <main+37>:   mov    DWORD PTR [esp],0x8048500
0x08048420 <main+44>:   call   0x804832c <puts@plt>
0x08048425 <main+49>:   jmp    0x8048433 <main+63>
0x08048427 <main+51>:   mov    DWORD PTR [esp],0x8048529
0x0804842e <main+58>:   call   0x804832c <puts@plt>
0x08048433 <main+63>:   leave
0x08048434 <main+64>:   ret

I understand that when it says sub esp,60 it's making a stack frame for the Main function. So why does it initialize the modified variable(mov DWORD PTR [esp+0x5c],0x0) at 5c in the stack frame and not at the bottom? Also, why does it only make room for 60 items(sub esp,60) when it knows there will be set length of 64?

1 Answer 1


Because it's 0x60 i.e. 96 in decimal. So it actually allocates 64 bytes for the buffer, then 4 bytes for modified. And the rest is 0x1С, which compiler added as a spare in the debug build.

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