manifest/revlog: do not let the revlog cache mutable objects
If a buffer of an mutable object is passed to revlog.addrevision(), the revlog
will happily store it in its cache. Later when the revlog reuses the cached
entry, if the manifest modified the object in-between, all kind of bugs
appears.
We fix it by:
- passing immutable objects to addrevision() if they are already available
- only storing the text in the cache if it's of str type
Then we can remove the conversion of the cache entry to str() during
retrieval. That was probably just there hiding the bug for the common cases
but not really fixing it.
# 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, incorporated herein by reference.
import heapq
def ancestor(a, b, pfunc):
"""
return the least common ancestor of nodes a and b or None if there
is no such ancestor.
pfunc must return a list of parent vertices
"""
if a == b:
return a
# find depth from root of all ancestors
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:
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 1:
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