remotefilelog: remove support for fetching packs
We've never used this at Google, and it's strictly an optimization
that indygreg and I are confident we can handle more elegantly
later. Let's rip it out to keep the surface area of the imported
remotefilelog small.
Some tests have changes because the tests for regular packs and
repacking were mixed with tests of fetching packs. I'm pretty sure
I've done the right thing here.
Differential Revision: https://phab.mercurial-scm.org/D5122
# 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)