revlog.revision: avoid opening the datafile without need.
If there's no inline data, revlog.revision opens the data file every
time it's called. This is useful if we're going to call chunk many
times, but, if we're going to call it only once, it's better to let
chunk open the file - if we're lucky, all the data we're going to need
is already cached and we won't need to even look at the file.
# changelog bisection for mercurial
#
# Copyright 2007 Matt Mackall
# Copyright 2005, 2006 Benoit Boissinot <benoit.boissinot@ens-lyon.org>
# Inspired by git bisect, extension skeleton taken from mq.py.
#
# This software may be used and distributed according to the terms
# of the GNU General Public License, incorporated herein by reference.
from i18n import _
import hg, util
def bisect(changelog, state):
clparents = changelog.parentrevs
skip = dict.fromkeys([changelog.rev(n) for n in state['skip']])
def buildancestors(bad, good):
# only the earliest bad revision matters
badrev = min([changelog.rev(n) for n in bad])
goodrevs = [changelog.rev(n) for n in good]
# build ancestors array
ancestors = [[]] * (changelog.count() + 1) # an extra for [-1]
# clear good revs from array
for node in goodrevs:
ancestors[node] = None
for rev in xrange(changelog.count(), -1, -1):
if ancestors[rev] is None:
for prev in clparents(rev):
ancestors[prev] = None
if ancestors[badrev] is None:
return badrev, None
return badrev, ancestors
good = 0
badrev, ancestors = buildancestors(state['bad'], state['good'])
if not ancestors: # looking for bad to good transition?
good = 1
badrev, ancestors = buildancestors(state['good'], state['bad'])
bad = changelog.node(badrev)
if not ancestors: # now we're confused
raise util.Abort(_("Inconsistent state, %s:%s is good and bad")
% (badrev, hg.short(bad)))
# build children dict
children = {}
visit = [badrev]
candidates = []
while visit:
rev = visit.pop(0)
if ancestors[rev] == []:
candidates.append(rev)
for prev in clparents(rev):
if prev != -1:
if prev in children:
children[prev].append(rev)
else:
children[prev] = [rev]
visit.append(prev)
candidates.sort()
# have we narrowed it down to one entry?
tot = len(candidates)
if tot == 1:
return (bad, 0, good)
perfect = tot / 2
# find the best node to test
best_rev = None
best_len = -1
poison = {}
for rev in candidates:
if rev in poison:
for c in children.get(rev, []):
poison[c] = True # poison children
continue
a = ancestors[rev] or [rev]
ancestors[rev] = None
x = len(a) # number of ancestors
y = tot - x # number of non-ancestors
value = min(x, y) # how good is this test?
if value > best_len and rev not in skip:
best_len = value
best_rev = rev
if value == perfect: # found a perfect candidate? quit early
break
if y < perfect: # all downhill from here?
for c in children.get(rev, []):
poison[c] = True # poison children
continue
for c in children.get(rev, []):
if ancestors[c]:
ancestors[c] = dict.fromkeys(ancestors[c] + a).keys()
else:
ancestors[c] = a + [c]
assert best_rev is not None
best_node = changelog.node(best_rev)
return (best_node, tot, good)