Mercurial > hg
view mercurial/ancestor.py @ 13436:b391c0c9be61 stable
hgwebdir: reduce memory usage for index generation
The archive list generator was holding a reference to each
temporary ui copy passed by rawentries(), so the memory
usage for index generation growed proportionally to the
ui object size and the amount of repositories. By returning a
list instead, the temporary reference is dropped immediately.
author | Wagner Bruna <wbruna@softwareexpress.com.br> |
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date | Thu, 17 Feb 2011 18:05:27 -0200 |
parents | 4cdaf1adafc8 |
children | 22565ddb28e7 |
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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