~bzr-pqm/bzr/bzr.dev

# (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

# TODO: Some kind of command-line display of revision properties: 

# perhaps show them in log -v and allow them as options to the commit command.

import bzrlib.errors

from bzrlib.graph import node_distances, select_farthest, all_descendants

from bzrlib.osutils import contains_whitespace

NULL_REVISION="null:"

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.

    parent_ids

        List of parent revision_ids

    properties

        Dictionary of revision properties.  These are attached to the

        revision as extra metadata.  The name must be a single 

        word; the value can be an arbitrary string.

    """

    def __init__(self, revision_id, properties=None, **args):

        self.revision_id = revision_id

        self.properties = properties or {}

        self._check_properties()

        self.parent_ids = []

        self.parent_sha1s = []

        self.__dict__.update(args)

    def __repr__(self):

        return "<Revision id %s>" % self.revision_id

    def __eq__(self, other):

        if not isinstance(other, Revision):

            return False

        # FIXME: rbc 20050930 parent_ids are not being compared

        return (

                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.properties == other.properties)

    def __ne__(self, other):

        return not self.__eq__(other)

    def _check_properties(self):

        """Verify that all revision properties are OK.

        """

        for name, value in self.properties.iteritems():

            if not isinstance(name, basestring) or contains_whitespace(name):

                raise ValueError("invalid property name %r" % name)

            if not isinstance(value, basestring):

                raise ValueError("invalid property value %r for %r" % 

                                 (name, value))

def is_ancestor(revision_id, candidate_id, branch):

    """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.

    """

    return candidate_id in branch.get_ancestry(revision_id)

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(revision.parent_ids)

        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.

    RBC: 20051024: note that when we have two partial histories, this may not

         be possible. But if we are willing to pretend :)... sure.

    """

    ancestors = {}

    descendants = {}

    lines = [revision]

    root = None

    descendants[revision] = {}

    while len(lines) > 0:

        new_lines = set()

        for line in lines:

            if line == NULL_REVISION:

                parents = []

                root = NULL_REVISION

            else:

                try:

                    rev = revision_source.get_revision(line)

                    parents = list(rev.parent_ids)

                    if len(parents) == 0:

                        parents = [NULL_REVISION]

                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

    if root is None:

        # The history for revision becomes inaccessible without

        # actually hitting a no-parents revision. This then

        # makes these asserts below trigger. So, if root is None

        # determine the actual root by walking the accessible tree

        # and then stash NULL_REVISION at the end.

        root = NULL_REVISION

        descendants[root] = {}

        # for every revision, check we can access at least

        # one parent, if we cant, add NULL_REVISION and

        # a link

        for rev in ancestors:

            if len(ancestors[rev]) == 0:

                raise RuntimeError('unreachable code ?!')

            ok = False

            for parent in ancestors[rev]:

                if parent in ancestors:

                    ok = True

            if ok:

                continue

            descendants[root][rev] = 1

            ancestors[rev].add(root)

        ancestors[root] = set()

    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)

    if root != root_b:

        raise bzrlib.errors.NoCommonRoot(revision_a, revision_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] = {}

        descendants[node].update(node_dec)

    return root, ancestors, descendants, common

def common_ancestor(revision_a, revision_b, revision_source):

    try:

        root, ancestors, descendants, common = \

            combined_graph(revision_a, revision_b, revision_source)

    except bzrlib.errors.NoCommonRoot:

        raise bzrlib.errors.NoCommonAncestor(revision_a, revision_b)

    distances = node_distances (descendants, ancestors, root)

    farthest = select_farthest(distances, common)

    if farthest is None or farthest == NULL_REVISION:

        raise bzrlib.errors.NoCommonAncestor(revision_a, revision_b)

    return farthest

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