##// END OF EJS Templates
cleanup: join string literals that are already on one line...
cleanup: join string literals that are already on one line Thanks to Kyle for noticing this and for providing the regular expression to run on the codebase. This patch has been reviewed by the test suite and they approved of it. # skip-blame: fallout from mass reformatting Differential Revision: https://phab.mercurial-scm.org/D7028

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bdiff.py
84 lines | 2.3 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(b"struct bdiff_line**")
b = ffi.new(b"struct bdiff_line**")
ac = ffi.new(b"char[]", str(sa))
bc = ffi.new(b"char[]", str(sb))
l = ffi.new(b"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(b"struct bdiff_line**")
b = ffi.new(b"struct bdiff_line**")
ac = ffi.new(b"char[]", str(sa))
bc = ffi.new(b"char[]", str(sb))
l = ffi.new(b"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(
b">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 b"".join(rl)