# Copyright (C) 2007-2009 by the Free Software Foundation, Inc.
#
# This file is part of GNU Mailman.
#
# GNU Mailman is free software: you can redistribute it and/or modify it under
# the terms of the GNU General Public License as published by the Free
# Software Foundation, either version 3 of the License, or (at your option)
# any later version.
#
# GNU Mailman is distributed in the hope that it will be useful, but WITHOUT
# ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or
# FITNESS FOR A PARTICULAR PURPOSE.  See the GNU General Public License for
# more details.
#
# You should have received a copy of the GNU General Public License along with
# GNU Mailman.  If not, see <http://www.gnu.org/licenses/>.

"""Password hashing and verification schemes.

Represents passwords using RFC 2307 syntax (as best we can tell).
"""

import os
import re
import hmac
import hashlib

from array import array
from base64 import urlsafe_b64decode as decode
from base64 import urlsafe_b64encode as encode
from munepy import Enum

from mailman.core import errors

SALT_LENGTH = 20 # bytes
ITERATIONS  = 2000

__all__ = [
    'Schemes',
    'make_secret',
    'check_response',
    ]



class PasswordScheme(object):
    TAG = ''

    @staticmethod
    def make_secret(password):
        """Return the hashed password"""
        raise NotImplementedError

    @staticmethod
    def check_response(challenge, response):
        """Return True if response matches challenge.

        It is expected that the scheme specifier prefix is already stripped
        from the response string.
        """
        raise NotImplementedError



class NoPasswordScheme(PasswordScheme):
    TAG = 'NONE'

    @staticmethod
    def make_secret(password):
        return ''

    @staticmethod
    def check_response(challenge, response):
        return False



class ClearTextPasswordScheme(PasswordScheme):
    TAG = 'CLEARTEXT'

    @staticmethod
    def make_secret(password):
        return password

    @staticmethod
    def check_response(challenge, response):
        return challenge == response



class SHAPasswordScheme(PasswordScheme):
    TAG = 'SHA'

    @staticmethod
    def make_secret(password):
        h = hashlib.sha1(password)
        return encode(h.digest())

    @staticmethod
    def check_response(challenge, response):
        h = hashlib.sha1(response)
        return challenge == encode(h.digest())



class SSHAPasswordScheme(PasswordScheme):
    TAG = 'SSHA'

    @staticmethod
    def make_secret(password):
        salt = os.urandom(SALT_LENGTH)
        h = hashlib.sha1(password)
        h.update(salt)
        return encode(h.digest() + salt)

    @staticmethod
    def check_response(challenge, response):
        # Get the salt from the challenge
        challenge_bytes = decode(challenge)
        digest = challenge_bytes[:20]
        salt = challenge_bytes[20:]
        h = hashlib.sha1(response)
        h.update(salt)
        return digest == h.digest()



# Basic algorithm given by Bob Fleck
class PBKDF2PasswordScheme(PasswordScheme):
    # This is a bit nasty if we wanted a different prf or iterations.  OTOH,
    # we really have no clue what the standard LDAP-ish specification for
    # those options is.
    TAG = 'PBKDF2 SHA %d' % ITERATIONS

    @staticmethod
    def _pbkdf2(password, salt, iterations):
        """From RFC2898 sec. 5.2.  Simplified to handle only 20 byte output
        case.  Output of 20 bytes means always exactly one block to handle,
        and a constant block counter appended to the salt in the initial hmac
        update.
        """
        h = hmac.new(password, None, hashlib.sha1)
        prf = h.copy()
        prf.update(salt + '\x00\x00\x00\x01')
        T = U = array('l', prf.digest())
        while iterations:
            prf = h.copy()
            prf.update(U.tostring())
            U = array('l', prf.digest())
            T = array('l', (t ^ u for t, u in zip(T, U)))
            iterations -= 1
        return T.tostring()

    @staticmethod
    def make_secret(password):
        """From RFC2898 sec. 5.2.  Simplified to handle only 20 byte output
        case.  Output of 20 bytes means always exactly one block to handle,
        and a constant block counter appended to the salt in the initial hmac
        update.
        """
        salt = os.urandom(SALT_LENGTH)
        digest = PBKDF2PasswordScheme._pbkdf2(password, salt, ITERATIONS)
        derived_key = encode(digest + salt)
        return derived_key

    @staticmethod
    def check_response(challenge, response, prf, iterations):
        # Decode the challenge to get the number of iterations and salt
        # XXX we don't support anything but sha prf
        if prf.lower() <> 'sha':
            return False
        try:
            iterations = int(iterations)
        except (ValueError, TypeError):
            return False
        challenge_bytes = decode(challenge)
        digest = challenge_bytes[:20]
        salt = challenge_bytes[20:]
        key = PBKDF2PasswordScheme._pbkdf2(response, salt, iterations)
        return digest == key



class Schemes(Enum):
    # no_scheme is deliberately ugly because no one should be using it.  Yes,
    # this makes cleartext inconsistent, but that's a common enough
    # terminology to justify the missing underscore.
    no_scheme   = 1
    cleartext   = 2
    sha         = 3
    ssha        = 4
    pbkdf2      = 5


_SCHEMES_BY_ENUM = {
    Schemes.no_scheme   : NoPasswordScheme,
    Schemes.cleartext   : ClearTextPasswordScheme,
    Schemes.sha         : SHAPasswordScheme,
    Schemes.ssha        : SSHAPasswordScheme,
    Schemes.pbkdf2      : PBKDF2PasswordScheme,
    }


# Some scheme tags have arguments, but the key for this dictionary should just
# be the lowercased scheme name.
_SCHEMES_BY_TAG = dict((_SCHEMES_BY_ENUM[e].TAG.split(' ')[0].lower(), e)
                       for e in _SCHEMES_BY_ENUM)

_DEFAULT_SCHEME = NoPasswordScheme



def make_secret(password, scheme=None):
    # The hash algorithms operate on bytes not strings.  The password argument
    # as provided here by the client will be a string (in Python 2 either
    # unicode or 8-bit, in Python 3 always unicode).  We need to encode this
    # string into a byte array, and the way to spell that in Python 2 is to
    # encode the string to utf-8.  The returned secret is a string, so it must
    # be a unicode.
    if isinstance(password, unicode):
        password = password.encode('utf-8')
    scheme_class = _SCHEMES_BY_ENUM.get(scheme)
    if not scheme_class:
        raise errors.BadPasswordSchemeError(scheme)
    secret = scheme_class.make_secret(password)
    return '{%s}%s' % (scheme_class.TAG, secret)


def check_response(challenge, response):
    mo = re.match(r'{(?P<scheme>[^}]+?)}(?P<rest>.*)',
                  challenge, re.IGNORECASE)
    if not mo:
        return False
    # See above for why we convert here.  However because we should have
    # generated the challenge, we assume that it is already a byte string.
    if isinstance(response, unicode):
        response = response.encode('utf-8')
    scheme_group, rest_group = mo.group('scheme', 'rest')
    scheme_parts = scheme_group.split()
    scheme       = scheme_parts[0].lower()
    scheme_enum  = _SCHEMES_BY_TAG.get(scheme, _DEFAULT_SCHEME)
    scheme_class = _SCHEMES_BY_ENUM[scheme_enum]
    if isinstance(rest_group, unicode):
        rest_group = rest_group.encode('utf-8')
    return scheme_class.check_response(rest_group, response, *scheme_parts[1:])


def lookup_scheme(scheme_name):
    return _SCHEMES_BY_TAG.get(scheme_name.lower())