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py3: add alternate output on Python 3...
py3: add alternate output on Python 3 This is basically the same deal as D5806. Python 3's exception printing output is different. We had to tweak the regular expression to match appropriately, hence the added line of output. Differential Revision: https://phab.mercurial-scm.org/D5807

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r33573:857876eb merge 4.3-rc stable
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bdiff.py
76 lines | 2.1 KiB | text/x-python | PythonLexer
# bdiff.py - CFFI implementation of bdiff.c
#
# Copyright 2016 Maciej Fijalkowski <fijall@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 struct
from ..pure.bdiff import *
from . import _bdiff
ffi = _bdiff.ffi
lib = _bdiff.lib
def blocks(sa, sb):
a = ffi.new("struct bdiff_line**")
b = ffi.new("struct bdiff_line**")
ac = ffi.new("char[]", str(sa))
bc = ffi.new("char[]", str(sb))
l = ffi.new("struct bdiff_hunk*")
try:
an = lib.bdiff_splitlines(ac, len(sa), a)
bn = lib.bdiff_splitlines(bc, len(sb), b)
if not a[0] or not b[0]:
raise MemoryError
count = lib.bdiff_diff(a[0], an, b[0], bn, l)
if count < 0:
raise MemoryError
rl = [None] * count
h = l.next
i = 0
while h:
rl[i] = (h.a1, h.a2, h.b1, h.b2)
h = h.next
i += 1
finally:
lib.free(a[0])
lib.free(b[0])
lib.bdiff_freehunks(l.next)
return rl
def bdiff(sa, sb):
a = ffi.new("struct bdiff_line**")
b = ffi.new("struct bdiff_line**")
ac = ffi.new("char[]", str(sa))
bc = ffi.new("char[]", str(sb))
l = ffi.new("struct bdiff_hunk*")
try:
an = lib.bdiff_splitlines(ac, len(sa), a)
bn = lib.bdiff_splitlines(bc, len(sb), b)
if not a[0] or not b[0]:
raise MemoryError
count = lib.bdiff_diff(a[0], an, b[0], bn, l)
if count < 0:
raise MemoryError
rl = []
h = l.next
la = lb = 0
while h:
if h.a1 != la or h.b1 != lb:
lgt = (b[0] + h.b1).l - (b[0] + lb).l
rl.append(struct.pack(">lll", (a[0] + la).l - a[0].l,
(a[0] + h.a1).l - a[0].l, lgt))
rl.append(str(ffi.buffer((b[0] + lb).l, lgt)))
la = h.a2
lb = h.b2
h = h.next
finally:
lib.free(a[0])
lib.free(b[0])
lib.bdiff_freehunks(l.next)
return "".join(rl)