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src/trezor/crypto: add trezor.crypto.base32

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Tomas Susanka 2018-05-11 11:50:41 +02:00 committed by Pavol Rusnak
parent 4de376acd6
commit bb57000449
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GPG Key ID: 91F3B339B9A02A3D
2 changed files with 143 additions and 0 deletions

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# Base32 implementation taken from the micropython-lib's base64 module
# https://github.com/micropython/micropython-lib/blob/master/base64/base64.py
#
from ubinascii import unhexlify
from ustruct import unpack
_b32alphabet = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ234567'
_b32tab = [ord(c) for c in _b32alphabet]
_b32rev = dict([(ord(v), k) for k, v in enumerate(_b32alphabet)])
def encode(s: bytes) -> str:
quanta, leftover = divmod(len(s), 5)
# Pad the last quantum with zero bits if necessary
if leftover:
s = s + bytes(5 - leftover) # Don't use += !
quanta += 1
encoded = bytearray()
for i in range(quanta):
# c1 and c2 are 16 bits wide, c3 is 8 bits wide. The intent of this
# code is to process the 40 bits in units of 5 bits. So we take the 1
# leftover bit of c1 and tack it onto c2. Then we take the 2 leftover
# bits of c2 and tack them onto c3. The shifts and masks are intended
# to give us values of exactly 5 bits in width.
c1, c2, c3 = unpack('!HHB', s[i*5:(i+1)*5])
c2 += (c1 & 1) << 16 # 17 bits wide
c3 += (c2 & 3) << 8 # 10 bits wide
encoded += bytes([_b32tab[c1 >> 11], # bits 1 - 5
_b32tab[(c1 >> 6) & 0x1f], # bits 6 - 10
_b32tab[(c1 >> 1) & 0x1f], # bits 11 - 15
_b32tab[c2 >> 12], # bits 16 - 20 (1 - 5)
_b32tab[(c2 >> 7) & 0x1f], # bits 21 - 25 (6 - 10)
_b32tab[(c2 >> 2) & 0x1f], # bits 26 - 30 (11 - 15)
_b32tab[c3 >> 5], # bits 31 - 35 (1 - 5)
_b32tab[c3 & 0x1f], # bits 36 - 40 (1 - 5)
])
# Adjust for any leftover partial quanta
if leftover == 1:
encoded = encoded[:-6] + b'======'
elif leftover == 2:
encoded = encoded[:-4] + b'===='
elif leftover == 3:
encoded = encoded[:-3] + b'==='
elif leftover == 4:
encoded = encoded[:-1] + b'='
return bytes(encoded).decode()
def decode(s: str) -> bytes:
s = s.encode()
quanta, leftover = divmod(len(s), 8)
if leftover:
raise ValueError('Incorrect padding')
# Strip off pad characters from the right. We need to count the pad
# characters because this will tell us how many null bytes to remove from
# the end of the decoded string.
padchars = s.find(b'=')
if padchars > 0:
padchars = len(s) - padchars
s = s[:-padchars]
else:
padchars = 0
# Now decode the full quanta
parts = []
acc = 0
shift = 35
for c in s:
val = _b32rev.get(c)
if val is None:
raise ValueError('Non-base32 digit found')
acc += _b32rev[c] << shift
shift -= 5
if shift < 0:
parts.append(unhexlify(('%010x' % acc).encode()))
acc = 0
shift = 35
# Process the last, partial quanta
last = unhexlify(bytes('%010x' % acc, "ascii"))
if padchars == 0:
last = b'' # No characters
elif padchars == 1:
last = last[:-1]
elif padchars == 3:
last = last[:-2]
elif padchars == 4:
last = last[:-3]
elif padchars == 6:
last = last[:-4]
else:
raise ValueError('Incorrect padding')
parts.append(last)
return b''.join(parts)

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from common import *
from trezor.crypto import base32
class TestCryptoBase32(unittest.TestCase):
# test vectors from:
# https://tools.ietf.org/html/rfc4648
# https://github.com/emn178/hi-base32/blob/master/tests/test.js
vectors = [
(b'', ''),
(b'f', 'MY======'),
(b'fo', 'MZXQ===='),
(b'foo', 'MZXW6==='),
(b'foob', 'MZXW6YQ='),
(b'fooba', 'MZXW6YTB'),
(b'foobar', 'MZXW6YTBOI======'),
(b'H', 'JA======'),
(b'He', 'JBSQ===='),
(b'Hel', 'JBSWY==='),
(b'Hell', 'JBSWY3A='),
(b'Hello', 'JBSWY3DP'),
(b'zlutoucky kun upel dabelske ody', 'PJWHK5DPOVRWW6JANN2W4IDVOBSWYIDEMFRGK3DTNNSSA33EPE======'),
(b'中文', '4S4K3ZUWQ4======'),
(b'中文1', '4S4K3ZUWQ4YQ===='),
(b'中文12', '4S4K3ZUWQ4YTE==='),
(b'aécio', 'MHB2SY3JN4======'),
(b'𠜎', '6CQJZDQ='),
(b'Base64是一種基於64個可列印字元來表示二進制資料的表示方法',
'IJQXGZJWGTTJRL7EXCAOPKFO4WP3VZUWXQ3DJZMARPSY7L7FRCL6LDNQ4WWZPZMFQPSL5BXIUGUOPJF24S5IZ2MAWLSYRNXIWOD6NFUZ46NIJ2FBVDT2JOXGS246NM4V')
]
def test_encode(self):
for a, b in self.vectors:
self.assertEqual(base32.encode(a), b)
def test_decode(self):
for a, b in self.vectors:
self.assertEqual(base32.decode(b), a)
if __name__ == '__main__':
unittest.main()