label: enforce the lack of leading or trailing white space
In practice, all commands create label are currently striping external white
space. Let us enforce this logic at a lower level before starting to rely on
it elsewhere.
# 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 + '\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 = bytes(a).splitlines(True)
b = bytes(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