merge: do early copy to deal with issue636
Without copies/renames, merges source names are 1:1 with their
targets. Copies and renames introduce the possibility that there will
be two merges with the same input but different output. By doing the
copy to the destination name before the merge, the actual merge
becomes 1:1 again, and no source is the input to two different merges.
- add a preliminary scan to applyupdates to do copies
- for the merge action, pass the old name (for finding ancestors) and
the new name (for input to the merge) to filemerge
- eliminate the old post-merge copy
- lookup file contents from new name in filemerge
- pass new name to external merge helper
- report merge failure at new name
- add a test
# 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, 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
visit = [a, b]
depth = {}
while visit:
vertex = visit[-1]
pl = pfunc(vertex)
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 = {}
while h:
d, n = heapq.heappop(h)
if n not in seen:
seen[n] = 1
yield (d, n)
for p in pfunc(n):
heapq.heappush(h, (depth[p], p))
def generations(vertex):
sg, s = None, {}
for g, v in ancestors(vertex):
if g != sg:
if sg:
yield sg, s
sg, s = g, {v:1}
else:
s[v] = 1
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