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hghave: avoid a deadlock reading the child process's output...
hghave: avoid a deadlock reading the child process's output The output of `pylint` is voluminous enough that it fills the buffer on Windows, and waited for the parent to read it out. But the parent was waiting on the child to exit. I'm not sure what the intent of setting `ret = -1` in the exception handler just above this was...

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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)