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revlog: move the details of revlog "v1" index inside revlog.utils.constants...
revlog: move the details of revlog "v1" index inside revlog.utils.constants The revlog module is quite large and this kind of format information would handy for other module. So let us start to gather this information about the format in a more appropriate place. We update various reference to this information to use the new "source of truth" in the process. Differential Revision: https://phab.mercurial-scm.org/D10304

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base85.py
88 lines | 2.0 KiB | text/x-python | PythonLexer
# 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