crypto.py 文件源码

python
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项目:CodingDojo 作者: ComputerSocietyUNB 项目源码 文件源码
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 2014, 100,000 iterations was the recommended default which took
        100ms on a 2.7Ghz Intel i7 with an optimized implementation. 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 about five times slower than OpenSSL's implementation. Look in
        django.contrib.auth.hashers for the present default, it is lower than
        the recommended 100,000 because of the performance difference between
        this and an optimized implementation.
        """
        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):
            u = salt + struct.pack(b'>I', i)
            result = 0
            for j in range(int(iterations)):
                dig1, dig2 = inner.copy(), outer.copy()
                dig1.update(u)
                dig2.update(dig1.digest())
                u = dig2.digest()
                result ^= _bin_to_long(u)
            return _long_to_bin(result, hex_format_string)

        T = [F(x) for x in range(1, l)]
        return b''.join(T) + F(l)[:r]
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