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revlogutils: remember known metadata parents for issue6528...
revlogutils: remember known metadata parents for issue6528 In the cases where the parent revs tell us for sure that the parent has metadata, remember this fact to avoid content recomputations later.

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bdiff.py
89 lines | 2.4 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 annotations
import struct
from typing import (
List,
Tuple,
)
from ..pure.bdiff import *
from . import _bdiff # pytype: disable=import-error
ffi = _bdiff.ffi
lib = _bdiff.lib
def blocks(sa: bytes, sb: bytes) -> List[Tuple[int, int, int, int]]:
a = ffi.new("struct bdiff_line**")
b = ffi.new("struct bdiff_line**")
ac = ffi.new("char[]", bytes(sa))
bc = ffi.new("char[]", bytes(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 = [(0, 0, 0, 0)] * 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: bytes, sb: bytes) -> bytes:
a = ffi.new("struct bdiff_line**")
b = ffi.new("struct bdiff_line**")
ac = ffi.new("char[]", bytes(sa))
bc = ffi.new("char[]", bytes(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(
b">lll",
(a[0] + la).l - a[0].l,
(a[0] + h.a1).l - a[0].l,
lgt,
)
)
rl.append(bytes(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)