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pathauditor: add a way to skip file system check...
pathauditor: add a way to skip file system check We need to be able to skip it when looking at data within the history. Doing them in all cases leads to buggy behavior like issue4749.

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bdiff.py
87 lines | 2.3 KiB | text/x-python | PythonLexer
# bdiff.py - Python implementation of bdiff.c
#
# Copyright 2009 Matt Mackall <mpm@selenic.com> and others
#
# This software may be used and distributed according to the terms of the
# GNU General Public License version 2 or any later version.
import struct, difflib, re
def splitnewlines(text):
'''like str.splitlines, but only split on newlines.'''
lines = [l + '\n' for l in text.split('\n')]
if lines:
if lines[-1] == '\n':
lines.pop()
else:
lines[-1] = lines[-1][:-1]
return lines
def _normalizeblocks(a, b, blocks):
prev = None
r = []
for curr in blocks:
if prev is None:
prev = curr
continue
shift = 0
a1, b1, l1 = prev
a1end = a1 + l1
b1end = b1 + l1
a2, b2, l2 = curr
a2end = a2 + l2
b2end = b2 + l2
if a1end == a2:
while (a1end + shift < a2end and
a[a1end + shift] == b[b1end + shift]):
shift += 1
elif b1end == b2:
while (b1end + shift < b2end and
a[a1end + shift] == b[b1end + shift]):
shift += 1
r.append((a1, b1, l1 + shift))
prev = a2 + shift, b2 + shift, l2 - shift
r.append(prev)
return r
def bdiff(a, b):
a = str(a).splitlines(True)
b = str(b).splitlines(True)
if not a:
s = "".join(b)
return s and (struct.pack(">lll", 0, 0, len(s)) + s)
bin = []
p = [0]
for i in a: p.append(p[-1] + len(i))
d = difflib.SequenceMatcher(None, a, b).get_matching_blocks()
d = _normalizeblocks(a, b, d)
la = 0
lb = 0
for am, bm, size in d:
s = "".join(b[lb:bm])
if am > la or s:
bin.append(struct.pack(">lll", p[la], p[am], len(s)) + s)
la = am + size
lb = bm + size
return "".join(bin)
def blocks(a, b):
an = splitnewlines(a)
bn = splitnewlines(b)
d = difflib.SequenceMatcher(None, an, bn).get_matching_blocks()
d = _normalizeblocks(an, bn, d)
return [(i, i + n, j, j + n) for (i, j, n) in d]
def fixws(text, allws):
if allws:
text = re.sub('[ \t\r]+', '', text)
else:
text = re.sub('[ \t\r]+', ' ', text)
text = text.replace(' \n', '\n')
return text