mirror of
https://github.com/Sneed-Group/Poodletooth-iLand
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227 lines
8.9 KiB
Python
227 lines
8.9 KiB
Python
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# -*- coding: utf-8 -*-
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#
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# Cipher/PKCS1-v1_5.py : PKCS#1 v1.5
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#
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# ===================================================================
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# The contents of this file are dedicated to the public domain. To
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# the extent that dedication to the public domain is not available,
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# everyone is granted a worldwide, perpetual, royalty-free,
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# non-exclusive license to exercise all rights associated with the
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# contents of this file for any purpose whatsoever.
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# No rights are reserved.
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#
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# THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND,
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# EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF
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# MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE AND
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# NONINFRINGEMENT. IN NO EVENT SHALL THE AUTHORS OR COPYRIGHT HOLDERS
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# BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER LIABILITY, WHETHER IN AN
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# ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM, OUT OF OR IN
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# CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE
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# SOFTWARE.
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# ===================================================================
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"""RSA encryption protocol according to PKCS#1 v1.5
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See RFC3447__ or the `original RSA Labs specification`__ .
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This scheme is more properly called ``RSAES-PKCS1-v1_5``.
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**If you are designing a new protocol, consider using the more robust PKCS#1 OAEP.**
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As an example, a sender may encrypt a message in this way:
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>>> from Crypto.Cipher import PKCS1_v1_5
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>>> from Crypto.PublicKey import RSA
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>>> from Crypto.Hash import SHA
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>>>
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>>> message = 'To be encrypted'
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>>> h = SHA.new(message)
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>>>
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>>> key = RSA.importKey(open('pubkey.der').read())
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>>> cipher = PKCS1_v1_5.new(key)
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>>> ciphertext = cipher.encrypt(message+h.digest())
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At the receiver side, decryption can be done using the private part of
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the RSA key:
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>>> From Crypto.Hash import SHA
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>>> from Crypto import Random
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>>>
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>>> key = RSA.importKey(open('privkey.der').read())
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>>>
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>>> dsize = SHA.digest_size
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>>> sentinel = Random.new().read(15+dsize) # Let's assume that average data length is 15
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>>>
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>>> cipher = PKCS1_v1_5.new(key)
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>>> message = cipher.decrypt(ciphertext, sentinel)
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>>>
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>>> digest = SHA.new(message[:-dsize]).digest()
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>>> if digest==message[-dsize:]: # Note how we DO NOT look for the sentinel
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>>> print "Encryption was correct."
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>>> else:
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>>> print "Encryption was not correct."
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:undocumented: __revision__, __package__
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.. __: http://www.ietf.org/rfc/rfc3447.txt
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.. __: http://www.rsa.com/rsalabs/node.asp?id=2125.
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"""
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__revision__ = "$Id$"
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__all__ = [ 'new', 'PKCS115_Cipher' ]
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from Crypto.Util.number import ceil_div
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from Crypto.Util.py3compat import *
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import Crypto.Util.number
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class PKCS115_Cipher:
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"""This cipher can perform PKCS#1 v1.5 RSA encryption or decryption."""
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def __init__(self, key):
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"""Initialize this PKCS#1 v1.5 cipher object.
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:Parameters:
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key : an RSA key object
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If a private half is given, both encryption and decryption are possible.
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If a public half is given, only encryption is possible.
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"""
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self._key = key
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def can_encrypt(self):
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"""Return True if this cipher object can be used for encryption."""
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return self._key.can_encrypt()
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def can_decrypt(self):
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"""Return True if this cipher object can be used for decryption."""
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return self._key.can_decrypt()
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def encrypt(self, message):
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"""Produce the PKCS#1 v1.5 encryption of a message.
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This function is named ``RSAES-PKCS1-V1_5-ENCRYPT``, and is specified in
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section 7.2.1 of RFC3447.
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For a complete example see `Crypto.Cipher.PKCS1_v1_5`.
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:Parameters:
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message : byte string
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The message to encrypt, also known as plaintext. It can be of
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variable length, but not longer than the RSA modulus (in bytes) minus 11.
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:Return: A byte string, the ciphertext in which the message is encrypted.
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It is as long as the RSA modulus (in bytes).
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:Raise ValueError:
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If the RSA key length is not sufficiently long to deal with the given
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message.
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"""
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# TODO: Verify the key is RSA
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randFunc = self._key._randfunc
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# See 7.2.1 in RFC3447
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modBits = Crypto.Util.number.size(self._key.n)
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k = ceil_div(modBits,8) # Convert from bits to bytes
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mLen = len(message)
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# Step 1
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if mLen > k-11:
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raise ValueError("Plaintext is too long.")
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# Step 2a
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class nonZeroRandByte:
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def __init__(self, rf): self.rf=rf
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def __call__(self, c):
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while bord(c)==0x00: c=self.rf(1)[0]
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return c
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ps = tobytes(map(nonZeroRandByte(randFunc), randFunc(k-mLen-3)))
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# Step 2b
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em = b('\x00\x02') + ps + bchr(0x00) + message
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# Step 3a (OS2IP), step 3b (RSAEP), part of step 3c (I2OSP)
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m = self._key.encrypt(em, 0)[0]
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# Complete step 3c (I2OSP)
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c = bchr(0x00)*(k-len(m)) + m
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return c
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def decrypt(self, ct, sentinel):
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"""Decrypt a PKCS#1 v1.5 ciphertext.
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This function is named ``RSAES-PKCS1-V1_5-DECRYPT``, and is specified in
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section 7.2.2 of RFC3447.
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For a complete example see `Crypto.Cipher.PKCS1_v1_5`.
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:Parameters:
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ct : byte string
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The ciphertext that contains the message to recover.
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sentinel : any type
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The object to return to indicate that an error was detected during decryption.
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:Return: A byte string. It is either the original message or the ``sentinel`` (in case of an error).
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:Raise ValueError:
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If the ciphertext length is incorrect
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:Raise TypeError:
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If the RSA key has no private half.
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:attention:
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You should **never** let the party who submitted the ciphertext know that
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this function returned the ``sentinel`` value.
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Armed with such knowledge (for a fair amount of carefully crafted but invalid ciphertexts),
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an attacker is able to recontruct the plaintext of any other encryption that were carried out
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with the same RSA public key (see `Bleichenbacher's`__ attack).
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In general, it should not be possible for the other party to distinguish
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whether processing at the server side failed because the value returned
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was a ``sentinel`` as opposed to a random, invalid message.
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In fact, the second option is not that unlikely: encryption done according to PKCS#1 v1.5
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embeds no good integrity check. There is roughly one chance
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in 2^16 for a random ciphertext to be returned as a valid message
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(although random looking).
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It is therefore advisabled to:
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1. Select as ``sentinel`` a value that resembles a plausable random, invalid message.
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2. Not report back an error as soon as you detect a ``sentinel`` value.
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Put differently, you should not explicitly check if the returned value is the ``sentinel`` or not.
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3. Cover all possible errors with a single, generic error indicator.
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4. Embed into the definition of ``message`` (at the protocol level) a digest (e.g. ``SHA-1``).
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It is recommended for it to be the rightmost part ``message``.
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5. Where possible, monitor the number of errors due to ciphertexts originating from the same party,
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and slow down the rate of the requests from such party (or even blacklist it altogether).
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**If you are designing a new protocol, consider using the more robust PKCS#1 OAEP.**
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.. __: http://www.bell-labs.com/user/bleichen/papers/pkcs.ps
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"""
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# TODO: Verify the key is RSA
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# See 7.2.1 in RFC3447
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modBits = Crypto.Util.number.size(self._key.n)
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k = ceil_div(modBits,8) # Convert from bits to bytes
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# Step 1
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if len(ct) != k:
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raise ValueError("Ciphertext with incorrect length.")
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# Step 2a (O2SIP), 2b (RSADP), and part of 2c (I2OSP)
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m = self._key.decrypt(ct)
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# Complete step 2c (I2OSP)
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em = bchr(0x00)*(k-len(m)) + m
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# Step 3
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sep = em.find(bchr(0x00),2)
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if not em.startswith(b('\x00\x02')) or sep<10:
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return sentinel
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# Step 4
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return em[sep+1:]
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def new(key):
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"""Return a cipher object `PKCS115_Cipher` that can be used to perform PKCS#1 v1.5 encryption or decryption.
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:Parameters:
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key : RSA key object
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The key to use to encrypt or decrypt the message. This is a `Crypto.PublicKey.RSA` object.
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Decryption is only possible if *key* is a private RSA key.
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"""
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return PKCS115_Cipher(key)
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