##// END OF EJS Templates
repair: use context manager for lock management...
repair: use context manager for lock management If repo.lock() raised inside of the try block, 'tr' would have been None in the finally block where it tries to release(). Modernize the syntax instead of just winching the lock out of the try block. I found several other instances of acquiring the lock inside of the 'try', but those finally blocks handle None references. I also started switching some trivial try/finally blocks to context managers, but didn't get them all because indenting over 3x for lock, wlock and transaction would have spilled over 80 characters. That got me wondering if there should be a repo.rwlock(), to handle locking and unlocking in the proper order. It also looks like py27 supports supports multiple context managers for a single 'with' statement. Should I hold off on the rest until py26 is dropped?

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diffhelpers.py
62 lines | 1.6 KiB | text/x-python | PythonLexer
# diffhelpers.py - pure Python implementation of diffhelpers.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
def addlines(fp, hunk, lena, lenb, a, b):
while True:
todoa = lena - len(a)
todob = lenb - len(b)
num = max(todoa, todob)
if num == 0:
break
for i in xrange(num):
s = fp.readline()
c = s[0]
if s == "\\ No newline at end of file\n":
fix_newline(hunk, a, b)
continue
if c == "\n":
# Some patches may be missing the control char
# on empty lines. Supply a leading space.
s = " \n"
hunk.append(s)
if c == "+":
b.append(s[1:])
elif c == "-":
a.append(s)
else:
b.append(s[1:])
a.append(s)
return 0
def fix_newline(hunk, a, b):
l = hunk[-1]
# tolerate CRLF in last line
if l.endswith('\r\n'):
hline = l[:-2]
else:
hline = l[:-1]
c = hline[0]
if c in " +":
b[-1] = hline[1:]
if c in " -":
a[-1] = hline
hunk[-1] = hline
return 0
def testhunk(a, b, bstart):
alen = len(a)
blen = len(b)
if alen > blen - bstart:
return -1
for i in xrange(alen):
if a[i][1:] != b[i + bstart]:
return -1
return 0