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# (C) 2005 Canonical
# This program is free software; you can redistribute it and/or modify
# it under the terms of the GNU General Public License as published by
# the Free Software Foundation; either version 2 of the License, or
# (at your option) any later version.
# This program is distributed in the hope that it will be useful,
# but WITHOUT ANY WARRANTY; without even the implied warranty of
# MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
# GNU General Public License for more details.
# You should have received a copy of the GNU General Public License
# along with this program; if not, write to the Free Software
# Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA 02111-1307 USA
import bzrlib.errors
from bzrlib.graph import farthest_nodes, node_distances, all_descendants
class RevisionReference(object):
"""
Reference to a stored revision.
Includes the revision_id and revision_sha1.
"""
def __eq__(self, other):
try:
return self.revision_id == other.revision_id and \
self.revision_sha1 == other.revision_sha1
except AttributeError:
return False
def __init__(self, revision_id, revision_sha1=None):
self.revision_id = None
self.revision_sha1 = None
if revision_id == None \
or isinstance(revision_id, basestring):
self.revision_id = revision_id
else:
raise ValueError('bad revision_id %r' % revision_id)
if revision_sha1 != None:
if isinstance(revision_sha1, basestring) \
and len(revision_sha1) == 40:
self.revision_sha1 = revision_sha1
else:
raise ValueError('bad revision_sha1 %r' % revision_sha1)
class Revision(object):
"""Single revision on a branch.
Revisions may know their revision_hash, but only once they've been
written out. This is not stored because you cannot write the hash
into the file it describes.
After bzr 0.0.5 revisions are allowed to have multiple parents.
parents
List of parent revisions, each is a RevisionReference.
"""
def __init__(self, inventory_id=None, inventory_sha1=None,
revision_id=None, timestamp=None,
message=None, timezone=None,
committer=None, parents=None):
self.inventory_id = inventory_id
self.inventory_sha1 = inventory_sha1
self.revision_id = revision_id
self.timestamp = timestamp
self.message = message
self.timezone = timezone
self.committer = committer
if parents is not None:
self.parents = parents
else:
self.parents = []
def __eq__(self, other):
try:
return self.inventory_id == other.inventory_id and \
self.inventory_sha1 == other.inventory_sha1 and \
self.revision_id == other.revision_id and \
self.timestamp == other.timestamp and \
self.message == other.message and \
self.timezone == other.timezone and \
self.committer == other.committer and \
self.parents == other.parents
except AttributeError:
return False
def __repr__(self):
return "<Revision id %s>" % self.revision_id
def __eq__(self, other):
if not isinstance(other, Revision):
return False
return (self.inventory_id == other.inventory_id
and self.inventory_sha1 == other.inventory_sha1
and self.revision_id == other.revision_id
and self.timestamp == other.timestamp
and self.message == other.message
and self.timezone == other.timezone
and self.committer == other.committer)
def __ne__(self, other):
return not self.__eq__(other)
REVISION_ID_RE = None
def validate_revision_id(rid):
"""Check rid is syntactically valid for a revision id."""
global REVISION_ID_RE
if not REVISION_ID_RE:
import re
REVISION_ID_RE = re.compile('[\w:.-]+@[\w%.-]+--?[\w]+--?[0-9a-f]+\Z')
if not REVISION_ID_RE.match(rid):
raise ValueError("malformed revision-id %r" % rid)
def is_ancestor(revision_id, candidate_id, revision_source):
"""Return true if candidate_id is an ancestor of revision_id.
A false negative will be returned if any intermediate descendent of
candidate_id is not present in any of the revision_sources.
revisions_source is an object supporting a get_revision operation that
behaves like Branch's.
"""
for ancestor_id, distance in iter_ancestors(revision_id, revision_source):
if ancestor_id == candidate_id:
return True
return False
def iter_ancestors(revision_id, revision_source, only_present=False):
ancestors = (revision_id,)
distance = 0
while len(ancestors) > 0:
new_ancestors = []
for ancestor in ancestors:
if not only_present:
yield ancestor, distance
try:
revision = revision_source.get_revision(ancestor)
except bzrlib.errors.NoSuchRevision, e:
if e.revision == revision_id:
raise
else:
continue
if only_present:
yield ancestor, distance
new_ancestors.extend([p.revision_id for p in revision.parents])
ancestors = new_ancestors
distance += 1
def find_present_ancestors(revision_id, revision_source):
"""Return the ancestors of a revision present in a branch.
It's possible that a branch won't have the complete ancestry of
one of its revisions.
"""
found_ancestors = {}
anc_iter = enumerate(iter_ancestors(revision_id, revision_source,
only_present=True))
for anc_order, (anc_id, anc_distance) in anc_iter:
if not found_ancestors.has_key(anc_id):
found_ancestors[anc_id] = (anc_order, anc_distance)
return found_ancestors
def __get_closest(intersection):
intersection.sort()
matches = []
for entry in intersection:
if entry[0] == intersection[0][0]:
matches.append(entry[2])
return matches
def old_common_ancestor(revision_a, revision_b, revision_source):
"""Find the ancestor common to both revisions that is closest to both.
"""
from bzrlib.trace import mutter
a_ancestors = find_present_ancestors(revision_a, revision_source)
b_ancestors = find_present_ancestors(revision_b, revision_source)
a_intersection = []
b_intersection = []
# a_order is used as a tie-breaker when two equally-good bases are found
for revision, (a_order, a_distance) in a_ancestors.iteritems():
if b_ancestors.has_key(revision):
a_intersection.append((a_distance, a_order, revision))
b_intersection.append((b_ancestors[revision][1], a_order, revision))
mutter("a intersection: %r" % a_intersection)
mutter("b intersection: %r" % b_intersection)
a_closest = __get_closest(a_intersection)
if len(a_closest) == 0:
return None
b_closest = __get_closest(b_intersection)
assert len(b_closest) != 0
mutter ("a_closest %r" % a_closest)
mutter ("b_closest %r" % b_closest)
if a_closest[0] in b_closest:
return a_closest[0]
elif b_closest[0] in a_closest:
return b_closest[0]
else:
raise bzrlib.errors.AmbiguousBase((a_closest[0], b_closest[0]))
return a_closest[0]
def revision_graph(revision, revision_source):
"""Produce a graph of the ancestry of the specified revision.
Return root, ancestors map, descendants map
TODO: Produce graphs with the NULL revision as root, so that we can find
a common even when trees are not branches don't represent a single line
of descent.
"""
ancestors = {}
descendants = {}
lines = [revision]
root = None
descendants[revision] = {}
while len(lines) > 0:
new_lines = set()
for line in lines:
try:
rev = revision_source.get_revision(line)
parents = [p.revision_id for p in rev.parents]
if len(parents) == 0:
root = line
except bzrlib.errors.NoSuchRevision:
if line == revision:
raise
parents = None
if parents is not None:
for parent in parents:
if parent not in ancestors:
new_lines.add(parent)
if parent not in descendants:
descendants[parent] = {}
descendants[parent][line] = 1
if parents is not None:
ancestors[line] = set(parents)
lines = new_lines
assert root not in descendants[root]
assert root not in ancestors[root]
return root, ancestors, descendants
def combined_graph(revision_a, revision_b, revision_source):
"""Produce a combined ancestry graph.
Return graph root, ancestors map, descendants map, set of common nodes"""
root, ancestors, descendants = revision_graph(revision_a, revision_source)
root_b, ancestors_b, descendants_b = revision_graph(revision_b,
revision_source)
assert root == root_b
common = set()
for node, node_anc in ancestors_b.iteritems():
if node in ancestors:
common.add(node)
else:
ancestors[node] = set()
ancestors[node].update(node_anc)
for node, node_dec in descendants_b.iteritems():
if node not in descendants:
descendants[node] = set()
descendants[node].update(node_dec)
return root, ancestors, descendants, common
def common_ancestor(revision_a, revision_b, revision_source):
root, ancestors, descendants, common = \
combined_graph(revision_a, revision_b, revision_source)
nodes = farthest_nodes(descendants, ancestors, root)
for node in nodes:
if node in common:
return node
class MultipleRevisionSources(object):
"""Proxy that looks in multiple branches for revisions."""
def __init__(self, *args):
object.__init__(self)
assert len(args) != 0
self._revision_sources = args
def get_revision(self, revision_id):
for source in self._revision_sources:
try:
return source.get_revision(revision_id)
except bzrlib.errors.NoSuchRevision, e:
pass
raise e
def get_intervening_revisions(ancestor_id, rev_id, rev_source,
revision_history=None):
"""Find the longest line of descent from maybe_ancestor to revision.
Revision history is followed where possible.
If ancestor_id == rev_id, list will be empty.
Otherwise, rev_id will be the last entry. ancestor_id will never appear.
If ancestor_id is not an ancestor, NotAncestor will be thrown
"""
root, ancestors, descendants = revision_graph(rev_id, rev_source)
if len(descendants) == 0:
raise NoSuchRevision(rev_source, rev_id)
if ancestor_id not in descendants:
rev_source.get_revision(ancestor_id)
raise bzrlib.errors.NotAncestor(rev_id, ancestor_id)
root_descendants = all_descendants(descendants, ancestor_id)
root_descendants.add(ancestor_id)
if rev_id not in root_descendants:
raise bzrlib.errors.NotAncestor(rev_id, ancestor_id)
distances = node_distances(descendants, ancestors, ancestor_id,
root_descendants=root_descendants)
def best_ancestor(rev_id):
best = None
for anc_id in ancestors[rev_id]:
try:
distance = distances[anc_id]
except KeyError:
continue
if revision_history is not None and anc_id in revision_history:
return anc_id
elif best is None or distance > best[1]:
best = (anc_id, distance)
return best[0]
next = rev_id
path = []
while next != ancestor_id:
path.append(next)
next = best_ancestor(next)
path.reverse()
return path
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