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behaves like Branch's.
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for ancestor_id, distance in iter_ancestors(revision_id, revision_source):
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if ancestor_id == candidate_id:
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def iter_ancestors(revision_id, revision_source, only_present=False):
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ancestors = (revision_id,)
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while len(ancestors) > 0:
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new_ancestors = []
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for ancestor in ancestors:
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if ancestor == candidate_id:
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yield ancestor, distance
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revision = revision_source.get_revision(ancestor)
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except bzrlib.errors.NoSuchRevision, e:
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yield ancestor, distance
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new_ancestors.extend([p.revision_id for p in revision.parents])
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ancestors = new_ancestors
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def find_present_ancestors(revision_id, revision_source):
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"""Return the ancestors of a revision present in a branch.
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It's possible that a branch won't have the complete ancestry of
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one of its revisions.
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anc_iter = enumerate(iter_ancestors(revision_id, revision_source,
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for anc_order, (anc_id, anc_distance) in anc_iter:
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if not found_ancestors.has_key(anc_id):
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found_ancestors[anc_id] = (anc_order, anc_distance)
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return found_ancestors
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class AmbiguousBase(bzrlib.errors.BzrError):
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def __init__(self, bases):
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msg = "The correct base is unclear, becase %s are all equally close" %\
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bzrlib.errors.BzrError.__init__(self, msg)
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def common_ancestor(revision_a, revision_b, revision_source):
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"""Find the ancestor common to both revisions that is closest to both.
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from bzrlib.trace import mutter
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a_ancestors = find_present_ancestors(revision_a, revision_source)
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b_ancestors = find_present_ancestors(revision_b, revision_source)
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# a_order is used as a tie-breaker when two equally-good bases are found
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for revision, (a_order, a_distance) in a_ancestors.iteritems():
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if b_ancestors.has_key(revision):
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a_intersection.append((a_distance, a_order, revision))
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b_intersection.append((b_ancestors[revision][1], a_order, revision))
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mutter("a intersection: %r" % a_intersection)
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mutter("b intersection: %r" % b_intersection)
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def get_closest(intersection):
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for entry in intersection:
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if entry[0] == intersection[0][0]:
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matches.append(entry[2])
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a_closest = get_closest(a_intersection)
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if len(a_closest) == 0:
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b_closest = get_closest(b_intersection)
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assert len(b_closest) != 0
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mutter ("a_closest %r" % a_closest)
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mutter ("b_closest %r" % b_closest)
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if a_closest[0] in b_closest:
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elif b_closest[0] in a_closest:
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raise AmbiguousBase((a_closest[0], b_closest[0]))
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class MultipleRevisionSources(object):
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def __init__(self, *args):