Mercurial > hg
view mercurial/ancestor.py @ 11791:00cde9bddbe4
repair: do not compress partial bundle if we do not keep it on disk
A partial bundle is created to temporarily save revisions > rev but
not descending from the node to strip, to be able to restore the
changesets after stripping the changelog.
Since this bundle is not kept after the strip operation, and is not
user-visible, it is not necessary and should be faster to avoid
compression.
author | Nicolas Dumazet <nicdumz.commits@gmail.com> |
---|---|
date | Thu, 12 Aug 2010 16:53:23 +0900 |
parents | 67bb9d78f05e |
children | 4f8067c94729 |
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# 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): """ return a minimal-distance ancestor of nodes a and b, or None if there is no such ancestor. Note that there can be several ancestors with the same (minimal) distance, and the one returned is arbitrary. 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 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