119 lines
3.3 KiB
Python
119 lines
3.3 KiB
Python
# SPDX-License-Identifier: MIT
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#
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# Copyright (c) 2017 PySecretHandshake contributors (see AUTHORS for more details)
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#
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# Permission is hereby granted, free of charge, to any person obtaining a copy
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# of this software and associated documentation files (the "Software"), to deal
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# in the Software without restriction, including without limitation the rights
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# to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
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# copies of the Software, and to permit persons to whom the Software is
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# furnished to do so, subject to the following conditions:
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#
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# The above copyright notice and this permission notice shall be included in all
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# copies or substantial portions of the Software.
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#
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# THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
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# IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
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# FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE
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# AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
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# LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM,
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# OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE
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# SOFTWARE.
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"""Utilities
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---------
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"""
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import struct
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from typing import Generator, Sequence, TypeVar
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NONCE_SIZE = 24
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MAX_NONCE = 8 * NONCE_SIZE
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T = TypeVar("T")
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def inc_nonce(nonce: bytes) -> bytes:
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"""Increment nonce"""
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num = bytes_to_long(nonce) + 1
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if num > 2**MAX_NONCE:
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num = 0
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bnum = long_to_bytes(num)
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bnum = b"\x00" * (NONCE_SIZE - len(bnum)) + bnum
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return bnum
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def split_chunks(seq: Sequence[T], n: int) -> Generator[Sequence[T], None, None]:
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"""Split sequence in equal-sized chunks.
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The last chunk is not padded."""
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while seq:
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yield seq[:n]
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seq = seq[n:]
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# Stolen from PyCypto (Public Domain)
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def b(s: str) -> bytes:
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"""Shorthand for ``s.encode("latin-1")``"""
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return s.encode("latin-1") # utf-8 would cause some side-effects we don't want
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def long_to_bytes(n: int, blocksize: int = 0) -> bytes:
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"""Convert a long integer to a byte string.
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If optional ``blocksize`` is given and greater than zero, pad the front of the byte string with binary zeros so
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that the length is a multiple of blocksize.
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"""
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# after much testing, this algorithm was deemed to be the fastest
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s = b("")
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pack = struct.pack
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while n > 0:
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s = pack(">I", n & 0xFFFFFFFF) + s
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n = n >> 32
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# strip off leading zeros
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for i, c in enumerate(s):
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if c != b("\000")[0]:
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break
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else:
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# only happens when n == 0
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s = b("\000")
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i = 0
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s = s[i:]
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# add back some pad bytes. this could be done more efficiently w.r.t. the
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# de-padding being done above, but sigh...
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if blocksize > 0 and len(s) % blocksize:
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s = (blocksize - len(s) % blocksize) * b("\000") + s
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return s
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def bytes_to_long(s: bytes) -> int:
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"""Convert a byte string to a long integer.
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This is (essentially) the inverse of ``long_to_bytes()``.
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"""
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acc = 0
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unpack = struct.unpack
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length = len(s)
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if length % 4:
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extra = 4 - length % 4
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s = b("\000") * extra + s
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length = length + extra
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for i in range(0, length, 4):
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acc = (acc << 32) + unpack(">I", s[i : i + 4])[0]
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return acc
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