I can not overflow buffer

Question

I have seen a buffer overflow code but I can not over flow it. Is there any gcc option to compile that? Or any wrong with that code.

The code is:

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

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

     if(argc == 1) {
          errx(1, "please specify an argument\n");
     }

     modified = 0;
     strcpy(buffer, argv[1]);

     if(modified == 0x61626364) {
            printf("you have correctly got the variable to the right value\n");
     } else {
            printf("Try again, you got 0x%08x\n", modified);
     }
}

and I am trying to run it this way:
perl -e 'print "A"x64 . "dcba"' | xargs ./main

Answer 1

You need to know

1. Know the stack memory layout and the address difference between the variable modified and buffer You can solve it by finding the offset between modified and buffer as (char *)&modified - (char *)buffer
2. Your machine endianess. I have used the stack overflow answer for this purpose

The linked demonstrates how to run the modified code that serves the purpose of determining the correct argument as well as stack smashing. The first Demo provides you with the argument that you can feed to your second Demo