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merge: check for conflicting actions irrespective of length of bids...
merge: check for conflicting actions irrespective of length of bids We should for whether bids contain a combination of actions which conflict with each other. Since right now we only have couple of such combination, and combinations also consist of two actions, we were checking for them only when length of bids is 2. Let's check that irrespective of the length of bids.

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bdiff.py
99 lines | 2.4 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.
from __future__ import absolute_import
import difflib
import re
import struct
def splitnewlines(text):
'''like str.splitlines, but only split on newlines.'''
lines = [l + b'\n' for l in text.split(b'\n')]
if lines:
if lines[-1] == b'\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 = bytes(a).splitlines(True)
b = bytes(b).splitlines(True)
if not a:
s = b"".join(b)
return s and (struct.pack(b">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 = b"".join(b[lb:bm])
if am > la or s:
bin.append(struct.pack(b">lll", p[la], p[am], len(s)) + s)
la = am + size
lb = bm + size
return b"".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(b'[ \t\r]+', b'', text)
else:
text = re.sub(b'[ \t\r]+', b' ', text)
text = text.replace(b' \n', b'\n')
return text