##// END OF EJS Templates
rebase: in --confirm option just abort if hit a conflict...
rebase: in --confirm option just abort if hit a conflict Before this patch, it was prompting the user in both cases 1) when there is no conflict 2) when there is at least one conflict. But for simplicity we can just abort if we hit a conflict and no need to prompt in that case. Differential Revision: https://phab.mercurial-scm.org/D3944

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r33573:857876eb merge 4.3-rc stable
r38701:b3d0c97a @81 default
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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)