Mercurial > hg
view mercurial/ancestor.py @ 9827:4fe9ca519637
mdiff: fix diff -b/B/w on mixed whitespace hunks (issue127)
Previous code was computing hunks then checking if these hunks could be ignored
when taking whitespace/blank-lines options in accounts. This approach is simple
but fails with hunks containing both whitespace and non-whitespace changes, the
whole hunk is emitted while it can be mostly made of whitespace. The new
version normalize the whitespaces before hunk generation, and test for
blank-lines afterwards.
author | Patrick Mezard <pmezard@gmail.com> |
---|---|
date | Wed, 11 Nov 2009 18:31:42 +0100 |
parents | 23429ebd3f9d |
children | 806e6b6cb8d8 25e572394f5c |
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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, 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