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parsers: fix refcount bug on corrupt index...
parsers: fix refcount bug on corrupt index When we encounter a corrupt index, we "fail" the init but our destructor still gets called. On some systems, this was causing us to attempt to decref a dangling to self->data.

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ancestor.py
91 lines | 2.6 KiB | text/x-python | PythonLexer
# ancestor.py - generic DAG ancestor algorithm for mercurial
#
# Copyright 2006 Matt Mackall <mpm@selenic.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 heapq
def ancestor(a, b, pfunc):
"""
Returns the common ancestor of a and b that is furthest from a
root (as measured by longest path) or None if no ancestor is
found. If there are multiple common ancestors at the same
distance, the first one found is returned.
pfunc must return a list of parent vertices for a given vertex
"""
if a == b:
return a
a, b = sorted([a, b])
# find depth from root of all ancestors
# depth is stored as a negative for heapq
parentcache = {}
visit = [a, b]
depth = {}
while visit:
vertex = visit[-1]
pl = pfunc(vertex)
parentcache[vertex] = pl
if not pl:
depth[vertex] = 0
visit.pop()
else:
for p in pl:
if p == a or p == b: # did we find a or b as a parent?
return p # we're done
if p not in depth:
visit.append(p)
if visit[-1] == vertex:
# -(maximum distance of parents + 1)
depth[vertex] = min([depth[p] for p in pl]) - 1
visit.pop()
# traverse ancestors in order of decreasing distance from root
def ancestors(vertex):
h = [(depth[vertex], vertex)]
seen = set()
while h:
d, n = heapq.heappop(h)
if n not in seen:
seen.add(n)
yield (d, n)
for p in parentcache[n]:
heapq.heappush(h, (depth[p], p))
def generations(vertex):
sg, s = None, set()
for g, v in ancestors(vertex):
if g != sg:
if sg:
yield sg, s
sg, s = g, set((v,))
else:
s.add(v)
yield sg, s
x = generations(a)
y = generations(b)
gx = x.next()
gy = y.next()
# increment each ancestor list until it is closer to root than
# the other, or they match
try:
while True:
if gx[0] == gy[0]:
for v in gx[1]:
if v in gy[1]:
return v
gy = y.next()
gx = x.next()
elif gx[0] > gy[0]:
gy = y.next()
else:
gx = x.next()
except StopIteration:
return None