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match: handle excludes using new differencematcher...
match: handle excludes using new differencematcher As I've said on earlier patches, I'm hoping to use more composition of simpler matchers instead of the single complex matcher we currently have. This extracts a first new matcher that composes two other matchers. It matches if the first matcher matches but the second does not. As such, we can use it for excludes, which this patch also does. We'll remove the now-unncessary code for excludes in the next patch.

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base85.py
77 lines | 1.9 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
_b85chars = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ" \
"abcdefghijklmnopqrstuvwxyz!#$%&()*+-;<=>?@^_`{|}~"
_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 += '\0' * (4 - r)
longs = len(text) >> 2
words = struct.unpack('>%dL' % (longs), text)
out = ''.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]
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('>%dL' % (len(out)), *out)
if cl:
out = out[:-(5 - cl)]
return out