##// END OF EJS Templates
bdiff: adjust criteria for getting optimal longest match in the A side middle...
bdiff: adjust criteria for getting optimal longest match in the A side middle We prefer matches closer to the middle to balance recursion, as introduced in f1ca249696ed. For ranges with uneven length, matches starting exactly in the middle should have preference. That will be optimal for matches of length 1. We will thus accept equality in the half check. For ranges with even length, half was ceil'ed when calculated but we got the preference for low matches from the 'less than half' check. To get the same result as before when we also accept equality, floor it. Without that, test-annotate.t would show some different (still correct but less optimal) results. This will change the heuristics. Tests shows a slightly different output - and sometimes slightly smaller bundles. The bundle size for 4.0 (hg bundle --base null -r 4.0 x.hg) happens to go from 22804885 to 22803824 bytes - an 0.005% reduction.

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py3kcompat.py
68 lines | 2.1 KiB | text/x-python | PythonLexer
# py3kcompat.py - compatibility definitions for running hg in py3k
#
# Copyright 2010 Renato Cunha <renatoc@gmail.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 builtins
import numbers
Number = numbers.Number
def bytesformatter(format, args):
'''Custom implementation of a formatter for bytestrings.
This function currently relies on the string formatter to do the
formatting and always returns bytes objects.
>>> bytesformatter(20, 10)
0
>>> bytesformatter('unicode %s, %s!', ('string', 'foo'))
b'unicode string, foo!'
>>> bytesformatter(b'test %s', 'me')
b'test me'
>>> bytesformatter('test %s', 'me')
b'test me'
>>> bytesformatter(b'test %s', b'me')
b'test me'
>>> bytesformatter('test %s', b'me')
b'test me'
>>> bytesformatter('test %d: %s', (1, b'result'))
b'test 1: result'
'''
# The current implementation just converts from bytes to unicode, do
# what's needed and then convert the results back to bytes.
# Another alternative is to use the Python C API implementation.
if isinstance(format, Number):
# If the fixer erroneously passes a number remainder operation to
# bytesformatter, we just return the correct operation
return format % args
if isinstance(format, bytes):
format = format.decode('utf-8', 'surrogateescape')
if isinstance(args, bytes):
args = args.decode('utf-8', 'surrogateescape')
if isinstance(args, tuple):
newargs = []
for arg in args:
if isinstance(arg, bytes):
arg = arg.decode('utf-8', 'surrogateescape')
newargs.append(arg)
args = tuple(newargs)
ret = format % args
return ret.encode('utf-8', 'surrogateescape')
builtins.bytesformatter = bytesformatter
origord = builtins.ord
def fakeord(char):
if isinstance(char, int):
return char
return origord(char)
builtins.ord = fakeord
if __name__ == '__main__':
import doctest
doctest.testmod()