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stream-clone: check is a compatible protocol can be found...
stream-clone: check is a compatible protocol can be found The previous code was explicitly checking if "v2" is listed in the "stream" bundle2 capability. The new code is checking is there is anything common between the versions supported client side and server side overlaps. This prepare the introduction of more stream version than "v2".

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bdiff.py
88 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.
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(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 = [(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(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)