192 lines
6.7 KiB
Python
192 lines
6.7 KiB
Python
"""
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Django's standard crypto functions and utilities.
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"""
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from __future__ import unicode_literals
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import hmac
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import struct
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import hashlib
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import binascii
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import time
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# Use the system PRNG if possible
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import random
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try:
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random = random.SystemRandom()
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using_sysrandom = True
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except NotImplementedError:
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import warnings
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warnings.warn('A secure pseudo-random number generator is not available '
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'on your system. Falling back to Mersenne Twister.')
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using_sysrandom = False
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from django.conf import settings
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from django.utils.encoding import force_bytes
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from django.utils import six
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from django.utils.six.moves import range
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def salted_hmac(key_salt, value, secret=None):
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"""
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Returns the HMAC-SHA1 of 'value', using a key generated from key_salt and a
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secret (which defaults to settings.SECRET_KEY).
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A different key_salt should be passed in for every application of HMAC.
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"""
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if secret is None:
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secret = settings.SECRET_KEY
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key_salt = force_bytes(key_salt)
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secret = force_bytes(secret)
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# We need to generate a derived key from our base key. We can do this by
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# passing the key_salt and our base key through a pseudo-random function and
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# SHA1 works nicely.
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key = hashlib.sha1(key_salt + secret).digest()
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# If len(key_salt + secret) > sha_constructor().block_size, the above
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# line is redundant and could be replaced by key = key_salt + secret, since
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# the hmac module does the same thing for keys longer than the block size.
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# However, we need to ensure that we *always* do this.
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return hmac.new(key, msg=force_bytes(value), digestmod=hashlib.sha1)
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def get_random_string(length=12,
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allowed_chars='abcdefghijklmnopqrstuvwxyz'
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'ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789'):
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"""
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Returns a securely generated random string.
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The default length of 12 with the a-z, A-Z, 0-9 character set returns
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a 71-bit value. log_2((26+26+10)^12) =~ 71 bits
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"""
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if not using_sysrandom:
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# This is ugly, and a hack, but it makes things better than
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# the alternative of predictability. This re-seeds the PRNG
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# using a value that is hard for an attacker to predict, every
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# time a random string is required. This may change the
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# properties of the chosen random sequence slightly, but this
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# is better than absolute predictability.
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random.seed(
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hashlib.sha256(
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("%s%s%s" % (
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random.getstate(),
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time.time(),
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settings.SECRET_KEY)).encode('utf-8')
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).digest())
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return ''.join(random.choice(allowed_chars) for i in range(length))
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if hasattr(hmac, "compare_digest"):
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# Prefer the stdlib implementation, when available.
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def constant_time_compare(val1, val2):
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return hmac.compare_digest(force_bytes(val1), force_bytes(val2))
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else:
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def constant_time_compare(val1, val2):
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"""
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Returns True if the two strings are equal, False otherwise.
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The time taken is independent of the number of characters that match.
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For the sake of simplicity, this function executes in constant time only
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when the two strings have the same length. It short-circuits when they
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have different lengths. Since Django only uses it to compare hashes of
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known expected length, this is acceptable.
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"""
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if len(val1) != len(val2):
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return False
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result = 0
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if six.PY3 and isinstance(val1, bytes) and isinstance(val2, bytes):
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for x, y in zip(val1, val2):
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result |= x ^ y
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else:
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for x, y in zip(val1, val2):
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result |= ord(x) ^ ord(y)
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return result == 0
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def _bin_to_long(x):
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"""
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Convert a binary string into a long integer
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This is a clever optimization for fast xor vector math
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"""
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return int(binascii.hexlify(x), 16)
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def _long_to_bin(x, hex_format_string):
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"""
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Convert a long integer into a binary string.
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hex_format_string is like "%020x" for padding 10 characters.
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"""
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return binascii.unhexlify((hex_format_string % x).encode('ascii'))
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if hasattr(hashlib, "pbkdf2_hmac"):
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def pbkdf2(password, salt, iterations, dklen=0, digest=None):
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"""
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Implements PBKDF2 with the same API as Django's existing
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implementation, using the stdlib.
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This is used in Python 2.7.8+ and 3.4+.
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"""
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if digest is None:
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digest = hashlib.sha256
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if not dklen:
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dklen = None
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password = force_bytes(password)
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salt = force_bytes(salt)
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return hashlib.pbkdf2_hmac(
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digest().name, password, salt, iterations, dklen)
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else:
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def pbkdf2(password, salt, iterations, dklen=0, digest=None):
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"""
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Implements PBKDF2 as defined in RFC 2898, section 5.2
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HMAC+SHA256 is used as the default pseudo random function.
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As of 2014, 100,000 iterations was the recommended default which took
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100ms on a 2.7Ghz Intel i7 with an optimized implementation. This is
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probably the bare minimum for security given 1000 iterations was
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recommended in 2001. This code is very well optimized for CPython and
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is about five times slower than OpenSSL's implementation. Look in
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django.contrib.auth.hashers for the present default, it is lower than
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the recommended 100,000 because of the performance difference between
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this and an optimized implementation.
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"""
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assert iterations > 0
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if not digest:
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digest = hashlib.sha256
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password = force_bytes(password)
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salt = force_bytes(salt)
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hlen = digest().digest_size
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if not dklen:
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dklen = hlen
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if dklen > (2 ** 32 - 1) * hlen:
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raise OverflowError('dklen too big')
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l = -(-dklen // hlen)
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r = dklen - (l - 1) * hlen
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hex_format_string = "%%0%ix" % (hlen * 2)
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inner, outer = digest(), digest()
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if len(password) > inner.block_size:
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password = digest(password).digest()
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password += b'\x00' * (inner.block_size - len(password))
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inner.update(password.translate(hmac.trans_36))
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outer.update(password.translate(hmac.trans_5C))
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def F(i):
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u = salt + struct.pack(b'>I', i)
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result = 0
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for j in range(int(iterations)):
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dig1, dig2 = inner.copy(), outer.copy()
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dig1.update(u)
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dig2.update(dig1.digest())
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u = dig2.digest()
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result ^= _bin_to_long(u)
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return _long_to_bin(result, hex_format_string)
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T = [F(x) for x in range(1, l)]
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return b''.join(T) + F(l)[:r]
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