Mercurial > hg
view mercurial/hbisect.py @ 6461:eb69e7989145
tests: easier error diagnostics for test-serve
author | Dirkjan Ochtman <dirkjan@ochtman.nl> |
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date | Thu, 03 Apr 2008 12:46:37 +0200 |
parents | fe8dbbe9520d |
children | 683428d1e639 8f256bf98219 |
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# 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 _ from node import short import 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, 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)