django1/django/utils/crypto.py

"""
Django's standard crypto functions and utilities.
"""

import hmac
import struct
import hashlib
import binascii
import operator
from django.conf import settings


_trans_5c = "".join([chr(x ^ 0x5C) for x in xrange(256)])
_trans_36 = "".join([chr(x ^ 0x36) for x in xrange(256)])


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).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=value, digestmod=hashlib.sha1)


def get_random_string(length=12, allowed_chars='abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789'):
    """
    Returns a random string of length characters from the set of a-z, A-Z, 0-9
    for use as a salt.

    The default length of 12 with the a-z, A-Z, 0-9 character set returns
    a 71-bit salt. log_2((26+26+10)^12) =~ 71 bits
    """
    import random
    try:
        random = random.SystemRandom()
    except NotImplementedError:
        pass
    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.
    """
    if len(val1) != len(val2):
        return False
    result = 0
    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 long(x.encode('hex'), 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)


def _fast_hmac(key, msg, digest):
    """
    A trimmed down version of Python's HMAC implementation
    """
    dig1, dig2 = digest(), digest()
    if len(key) > dig1.block_size:
        key = digest(key).digest()
    key += chr(0) * (dig1.block_size - len(key))
    dig1.update(key.translate(_trans_36))
    dig1.update(msg)
    dig2.update(key.translate(_trans_5c))
    dig2.update(dig1.digest())
    return dig2


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.

    Right now 10,000 iterations is the recommended default which takes
    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.
    """
    assert iterations > 0
    if not digest:
        digest = hashlib.sha256
    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)

    def F(i):
        def U():
            u = salt + struct.pack('>I', i)
            for j in xrange(int(iterations)):
                u = _fast_hmac(password, u, digest).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 ''.join(T[:-1]) + T[-1][:r]