""" Django's standard crypto functions and utilities. """ from __future__ import unicode_literals import hmac import struct import hashlib import binascii import operator import time from functools import reduce # Use the system PRNG if possible import random try: random = random.SystemRandom() using_sysrandom = True except NotImplementedError: import warnings warnings.warn('A secure pseudo-random number generator is not available ' 'on your system. Falling back to Mersenne Twister.') using_sysrandom = False from django.conf import settings from django.utils.encoding import force_bytes from django.utils import six from django.utils.six.moves import xrange def salted_hmac(key_salt, value, secret=None): """ Returns the HMAC-SHA1 of 'value', using a key generated from key_salt and a secret (which defaults to settings.SECRET_KEY). A different key_salt should be passed in for every application of HMAC. """ if secret is None: secret = settings.SECRET_KEY # We need to generate a derived key from our base key. We can do this by # passing the key_salt and our base key through a pseudo-random function and # SHA1 works nicely. key = hashlib.sha1((key_salt + secret).encode('utf-8')).digest() # If len(key_salt + secret) > sha_constructor().block_size, the above # line is redundant and could be replaced by key = key_salt + secret, since # the hmac module does the same thing for keys longer than the block size. # However, we need to ensure that we *always* do this. return hmac.new(key, msg=force_bytes(value), digestmod=hashlib.sha1) def get_random_string(length=12, allowed_chars='abcdefghijklmnopqrstuvwxyz' 'ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789'): """ Returns a securely generated random string. The default length of 12 with the a-z, A-Z, 0-9 character set returns a 71-bit value. log_2((26+26+10)^12) =~ 71 bits """ if not using_sysrandom: # This is ugly, and a hack, but it makes things better than # the alternative of predictability. This re-seeds the PRNG # using a value that is hard for an attacker to predict, every # time a random string is required. This may change the # properties of the chosen random sequence slightly, but this # is better than absolute predictability. random.seed( hashlib.sha256( ("%s%s%s" % ( random.getstate(), time.time(), settings.SECRET_KEY)).encode('utf-8') ).digest()) return ''.join(random.choice(allowed_chars) for i in range(length)) def constant_time_compare(val1, val2): """ Returns True if the two strings are equal, False otherwise. The time taken is independent of the number of characters that match. For the sake of simplicity, this function executes in constant time only when the two strings have the same length. It short-circuits when they have different lengths. Since Django only uses it to compare hashes of known expected length, this is acceptable. """ if len(val1) != len(val2): return False result = 0 if six.PY3 and isinstance(val1, bytes) and isinstance(val2, bytes): for x, y in zip(val1, val2): result |= x ^ y else: for x, y in zip(val1, val2): result |= ord(x) ^ ord(y) return result == 0 def _bin_to_long(x): """ Convert a binary string into a long integer This is a clever optimization for fast xor vector math """ return int(binascii.hexlify(x), 16) def _long_to_bin(x, hex_format_string): """ Convert a long integer into a binary string. hex_format_string is like "%020x" for padding 10 characters. """ return binascii.unhexlify((hex_format_string % x).encode('ascii')) def pbkdf2(password, salt, iterations, dklen=0, digest=None): """ Implements PBKDF2 as defined in RFC 2898, section 5.2 HMAC+SHA256 is used as the default pseudo random function. As of 2011, 10,000 iterations was the recommended default which took 100ms on a 2.2Ghz Core 2 Duo. This is probably the bare minimum for security given 1000 iterations was recommended in 2001. This code is very well optimized for CPython and is only four times slower than openssl's implementation. Look in django.contrib.auth.hashers for the present default. """ assert iterations > 0 if not digest: digest = hashlib.sha256 password = force_bytes(password) salt = force_bytes(salt) hlen = digest().digest_size if not dklen: dklen = hlen if dklen > (2 ** 32 - 1) * hlen: raise OverflowError('dklen too big') l = -(-dklen // hlen) r = dklen - (l - 1) * hlen hex_format_string = "%%0%ix" % (hlen * 2) inner, outer = digest(), digest() if len(password) > inner.block_size: password = digest(password).digest() password += b'\x00' * (inner.block_size - len(password)) inner.update(password.translate(hmac.trans_36)) outer.update(password.translate(hmac.trans_5C)) def F(i): def U(): u = salt + struct.pack(b'>I', i) for j in xrange(int(iterations)): dig1, dig2 = inner.copy(), outer.copy() dig1.update(u) dig2.update(dig1.digest()) u = dig2.digest() yield _bin_to_long(u) return _long_to_bin(reduce(operator.xor, U()), hex_format_string) T = [F(x) for x in range(1, l + 1)] return b''.join(T[:-1]) + T[-1][:r]