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# This program is free software; you can redistribute it and/or modify
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# it under the terms of the GNU General Public License as published by
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# the Free Software Foundation; either version 2 of the License, or
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# (at your option) any later version.
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# This program is distributed in the hope that it will be useful,
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# but WITHOUT ANY WARRANTY; without even the implied warranty of
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# MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
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# GNU General Public License for more details.
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# You should have received a copy of the GNU General Public License
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# along with this program; if not, write to the Free Software
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# Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA 02111-1307 USA
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from bzrlib.graph import node_distances, select_farthest, all_descendants
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class Revision(object):
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"""Single revision on a branch.
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Revisions may know their revision_hash, but only once they've been
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written out. This is not stored because you cannot write the hash
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into the file it describes.
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After bzr 0.0.5 revisions are allowed to have multiple parents.
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List of parent revision_ids
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def __init__(self, **args):
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self.__dict__.update(args)
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self.parent_sha1s = []
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return "<Revision id %s>" % self.revision_id
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def __eq__(self, other):
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if not isinstance(other, Revision):
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# FIXME: rbc 20050930 parent_ids are not being compared
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self.inventory_sha1 == other.inventory_sha1
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and self.revision_id == other.revision_id
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and self.timestamp == other.timestamp
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and self.message == other.message
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and self.timezone == other.timezone
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and self.committer == other.committer)
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def __ne__(self, other):
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return not self.__eq__(other)
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def validate_revision_id(rid):
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"""Check rid is syntactically valid for a revision id."""
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if not REVISION_ID_RE:
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REVISION_ID_RE = re.compile('[\w:.-]+@[\w%.-]+--?[\w]+--?[0-9a-f]+\Z')
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if not REVISION_ID_RE.match(rid):
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raise ValueError("malformed revision-id %r" % rid)
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def is_ancestor(revision_id, candidate_id, branch):
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"""Return true if candidate_id is an ancestor of revision_id.
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A false negative will be returned if any intermediate descendent of
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candidate_id is not present in any of the revision_sources.
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revisions_source is an object supporting a get_revision operation that
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behaves like Branch's.
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return candidate_id in branch.get_ancestry(revision_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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for ancestor in ancestors:
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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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if e.revision == revision_id:
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yield ancestor, distance
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new_ancestors.extend(revision.parent_ids)
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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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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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def old_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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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 bzrlib.errors.AmbiguousBase((a_closest[0], b_closest[0]))
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def revision_graph(revision, revision_source):
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"""Produce a graph of the ancestry of the specified revision.
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Return root, ancestors map, descendants map
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TODO: Produce graphs with the NULL revision as root, so that we can find
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a common even when trees are not branches don't represent a single line
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descendants[revision] = {}
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while len(lines) > 0:
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if line == NULL_REVISION:
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rev = revision_source.get_revision(line)
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parents = list(rev.parent_ids)
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if len(parents) == 0:
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parents = [NULL_REVISION]
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except bzrlib.errors.NoSuchRevision:
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if parents is not None:
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for parent in parents:
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if parent not in ancestors:
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new_lines.add(parent)
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if parent not in descendants:
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descendants[parent] = {}
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descendants[parent][line] = 1
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if parents is not None:
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ancestors[line] = set(parents)
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assert root not in descendants[root]
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assert root not in ancestors[root]
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return root, ancestors, descendants
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def combined_graph(revision_a, revision_b, revision_source):
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"""Produce a combined ancestry graph.
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Return graph root, ancestors map, descendants map, set of common nodes"""
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root, ancestors, descendants = revision_graph(revision_a, revision_source)
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root_b, ancestors_b, descendants_b = revision_graph(revision_b,
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raise bzrlib.errors.NoCommonRoot(revision_a, revision_b)
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for node, node_anc in ancestors_b.iteritems():
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if node in ancestors:
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ancestors[node] = set()
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ancestors[node].update(node_anc)
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for node, node_dec in descendants_b.iteritems():
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if node not in descendants:
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descendants[node] = {}
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descendants[node].update(node_dec)
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return root, ancestors, descendants, common
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def common_ancestor(revision_a, revision_b, revision_source):
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root, ancestors, descendants, common = \
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combined_graph(revision_a, revision_b, revision_source)
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except bzrlib.errors.NoCommonRoot:
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raise bzrlib.errors.NoCommonAncestor(revision_a, revision_b)
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distances = node_distances (descendants, ancestors, root)
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farthest = select_farthest(distances, common)
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if farthest is None or farthest == NULL_REVISION:
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raise bzrlib.errors.NoCommonAncestor(revision_a, revision_b)
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class MultipleRevisionSources(object):
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"""Proxy that looks in multiple branches for revisions."""
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def __init__(self, *args):
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object.__init__(self)
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assert len(args) != 0
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self._revision_sources = args
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def get_revision(self, revision_id):
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for source in self._revision_sources:
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return source.get_revision(revision_id)
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except bzrlib.errors.NoSuchRevision, e:
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def get_intervening_revisions(ancestor_id, rev_id, rev_source,
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revision_history=None):
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"""Find the longest line of descent from maybe_ancestor to revision.
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Revision history is followed where possible.
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If ancestor_id == rev_id, list will be empty.
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Otherwise, rev_id will be the last entry. ancestor_id will never appear.
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If ancestor_id is not an ancestor, NotAncestor will be thrown
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root, ancestors, descendants = revision_graph(rev_id, rev_source)
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if len(descendants) == 0:
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raise NoSuchRevision(rev_source, rev_id)
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if ancestor_id not in descendants:
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rev_source.get_revision(ancestor_id)
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raise bzrlib.errors.NotAncestor(rev_id, ancestor_id)
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root_descendants = all_descendants(descendants, ancestor_id)
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root_descendants.add(ancestor_id)
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if rev_id not in root_descendants:
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raise bzrlib.errors.NotAncestor(rev_id, ancestor_id)
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distances = node_distances(descendants, ancestors, ancestor_id,
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root_descendants=root_descendants)
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def best_ancestor(rev_id):
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for anc_id in ancestors[rev_id]:
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distance = distances[anc_id]
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if revision_history is not None and anc_id in revision_history:
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elif best is None or distance > best[1]:
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best = (anc_id, distance)
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while next != ancestor_id:
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next = best_ancestor(next)
added
removed
bzrlib/revision.py
