[Google CTF:sftp] Randomized heap :D 26 Jun 2018

[Google CTF:sftp] Randomized heap :D

The challenge.

This is the chall from google ctf named sftp. The task was relatively easy (60 solves), but I managed to fail it for the first time, and only with some hints from @kaanezder I got a clear understanding. Thanks, man!

This file server has a sophisticated malloc implementation designed to thwart traditional heap exploitation techniques…

sftp.ctfcompetition.com 1337


Let’s try it.

Actually, the description is already given us a very straight hint, but let’s just play with the file. The first thing is to guess password. In IDA we can see, that password is checked by hash:

  __printf_chk(1LL, "%s@%s's password: ", off_208B88);
  if ( !(unsigned int)__isoc99_scanf("%15s", &v5) )
    return 0LL;
  v3 = _IO_getc(stdin);
  LOWORD(v3) = v5;
  if ( !v5 )
    return 0LL;
  v4 = 0x5417;
    v3 ^= v4;
    v4 = 2 * v3;
    LOWORD(v3) = *v0;
  while ( (_BYTE)v3 );
  result = 1LL;
  if ( (_WORD)v4 != 0x8DFAu )
    return 0LL;
  return result; 

But the hash is only 2 bytes, so we can start brute force (if you do not find the origin password, there are huge chances to get collision):

#!/usr/bin/env python

from pwn import *
from struct import *

def hash(word):

	v4 = 0x5417
	v3 = 0

	for c in word:
		v3 = ord(c) ^ v4
		v4 = (v3 * 2) & 0xffff

		if v4 == 0x8DFA:
			log.info("got it! %s [%s]", word, c)

	log.info("%s %x", word, v4)

for ps in open("../../../rockyou.txt", 'r').readlines():

	#log.info("%s" % ps)

I’ve used rockyou dictionary, with only one exception: I check password after every character, not only after word. Ok, we found it, password = Steve (Corresponding entry in rockyou is Steven).

Also, you can notice, that you can input any symbols (0..255) not only printable ones.

Ok, now we got in, and we can explore bugs in binary. There are plenty of them. First I tried to create a lot of directories. Binary was crashing randomly after 100..10000 tries. Then I create a directory and make cd to it, in a loop, the crash happened much earlier. Then you can simply type cd / and got crash right after it. It was all different bugs, but briefly analyzing them, you can see, that all of them is not really useful. There are no leaks, there are no possibilities of RIP control (well, at least I couldn’t find any :D).

There is also sources file available as sftp:

Connected to sftp.google.ctf.
sftp> ls
sftp> ls src
sftp> get src/sftp.c
#include <stdbool.h>
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>


Wow, we can analyze at sources level now. Sources are obviously no full, but they do match our binary file.

Actual bug

Let’s remember task description, there was something about alloc(), right? Hm… binary is exporting some memory relative functions:

Beautiful! If you would break in gdb at rand() you can confirm that all memory allocation is just a fiction, pointers are randomly picked up from mapped memory region, never free, and not changed at reallocation.

Start              End                Perm	Name
0x40000000         0x60100000         rw-p	mapped

Now, the idea is to allocate memory, we will try to get overlapped chunks and make leaks from it. The size of mapped memory is not so small (>512MB). Btw, the binary is almost fully protected:

$ checksec sftp
    Arch:     amd64-64-little
    RELRO:    Partial RELRO
    Stack:    Canary found
    NX:       NX enabled
    PIE:      PIE enabled
    FORTIFY:  Enabled


Obviously, we need to allocate one (or small number) big chunk of memory and lots of small ones. The most suitable candidate for the big one is put file command, with maximum possible size:

#define file_max 65535

For the small chunks, I’ll use put file also, but only with 8 bytes (to match pointer size on x64 platform).

The idea is working, we can get a typical leak of binary structures:

This must be the structure entry and file_entry right after it.

struct entry {
	struct directory_entry* parent_directory;
	enum entry_type type;
	char name[name_max];

struct file_entry {
	struct entry entry;

	size_t size;
	char* data;

Now we can overwrite data with put command on the big chunk, then read or write small chunk to get or set any memory of process that we want. Let’s analyze the first pointer of the leaked structure. Actually, it points somewhere in .bss section. Since we put our files in root directory it must be the global variable root:

directory_entry* root = NULL;

Actually substructing 0x208be0 from this address we will get base of sftp image in memory. Now we can craft pointer to .got.plt section and read address to libc. Then we can overwrite pointer in .got.plt got get code execution. Briefly search of function candidates to overwrite, I saw that strrchr() is the perfect one, because it is called with the argument that we control. So we can simply write mkdir sh and strrchr() will be called with sh argument. We just need to change that pointer to get shell.

Here is my sploit ;-)