Mercurial > hg
view mercurial/pure/base85.py @ 46779:49fd21f32695
revlog: guarantee that p1 != null if a non-null parent exists
This change does not affect the hashing (which already did this
transformation), but can change the log output in the rare case where
this behavior was observed in repositories. The change can simplify
iteration code where regular changesets and merges are distinct
branches.
Differential Revision: https://phab.mercurial-scm.org/D10150
author | Joerg Sonnenberger <joerg@bec.de> |
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date | Wed, 10 Mar 2021 18:09:21 +0100 |
parents | 4cd911040ba5 |
children | 6000f5b25c9b |
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# base85.py: pure python base85 codec # # Copyright (C) 2009 Brendan Cully <brendan@kublai.com> # # This software may be used and distributed according to the terms of the # GNU General Public License version 2 or any later version. from __future__ import absolute_import import struct from .. import pycompat _b85chars = pycompat.bytestr( b"0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdef" b"ghijklmnopqrstuvwxyz!#$%&()*+-;<=>?@^_`{|}~" ) _b85chars2 = [(a + b) for a in _b85chars for b in _b85chars] _b85dec = {} def _mkb85dec(): for i, c in enumerate(_b85chars): _b85dec[c] = i def b85encode(text, pad=False): """encode text in base85 format""" l = len(text) r = l % 4 if r: text += b'\0' * (4 - r) longs = len(text) >> 2 words = struct.unpack(b'>%dL' % longs, text) out = b''.join( _b85chars[(word // 52200625) % 85] + _b85chars2[(word // 7225) % 7225] + _b85chars2[word % 7225] for word in words ) if pad: return out # Trim padding olen = l % 4 if olen: olen += 1 olen += l // 4 * 5 return out[:olen] def b85decode(text): """decode base85-encoded text""" if not _b85dec: _mkb85dec() l = len(text) out = [] for i in range(0, len(text), 5): chunk = text[i : i + 5] chunk = pycompat.bytestr(chunk) acc = 0 for j, c in enumerate(chunk): try: acc = acc * 85 + _b85dec[c] except KeyError: raise ValueError( 'bad base85 character at position %d' % (i + j) ) if acc > 4294967295: raise ValueError('Base85 overflow in hunk starting at byte %d' % i) out.append(acc) # Pad final chunk if necessary cl = l % 5 if cl: acc *= 85 ** (5 - cl) if cl > 1: acc += 0xFFFFFF >> (cl - 2) * 8 out[-1] = acc out = struct.pack(b'>%dL' % (len(out)), *out) if cl: out = out[: -(5 - cl)] return out