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Diffstat (limited to 'mailman/passwords.py')
| -rw-r--r-- | mailman/passwords.py | 253 |
1 files changed, 0 insertions, 253 deletions
diff --git a/mailman/passwords.py b/mailman/passwords.py deleted file mode 100644 index 0c8284c1c..000000000 --- a/mailman/passwords.py +++ /dev/null @@ -1,253 +0,0 @@ -# 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). -""" - -from __future__ import unicode_literals - -__metaclass__ = type -__all__ = [ - 'Schemes', - 'make_secret', - 'check_response', - ] - - -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 - - - -class PasswordScheme: - TAG = b'' - - @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 = b'NONE' - - @staticmethod - def make_secret(password): - return b'' - - @staticmethod - def check_response(challenge, response): - return False - - - -class ClearTextPasswordScheme(PasswordScheme): - TAG = b'CLEARTEXT' - - @staticmethod - def make_secret(password): - return password - - @staticmethod - def check_response(challenge, response): - return challenge == response - - - -class SHAPasswordScheme(PasswordScheme): - TAG = b'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 = b'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 = b'PBKDF2 SHA {0}'.format(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 + b'\x00\x00\x00\x01') - T = U = array(b'l', prf.digest()) - while iterations: - prf = h.copy() - prf.update(U.tostring()) - U = array(b'l', prf.digest()) - T = array(b'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() <> b'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 b'{{{0}}}{1}'.format(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()) |