78
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raise ValueError("invalid property value %r for %r" %
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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.repository.get_ancestry(revision_id)
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def get_history(self, repository):
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"""Return the canonical line-of-history for this revision.
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If ghosts are present this may differ in result from a ghost-free
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current_revision = self
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while current_revision is not None:
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reversed_result.append(current_revision.revision_id)
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if not len (current_revision.parent_ids):
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reversed_result.append(None)
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current_revision = None
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next_revision_id = current_revision.parent_ids[0]
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current_revision = repository.get_revision(next_revision_id)
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reversed_result.reverse()
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return reversed_result
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def get_summary(self):
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"""Get the first line of the log message for this revision.
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return self.message.lstrip().split('\n', 1)[0]
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def get_apparent_author(self):
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"""Return the apparent author of this revision.
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If the revision properties contain the author name,
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return it. Otherwise return the committer name.
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return self.properties.get('author', self.committer)
94
120
def iter_ancestors(revision_id, revision_source, only_present=False):
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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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def is_reserved_id(revision_id):
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"""Determine whether a revision id is reserved
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:return: True if the revision is is reserved, False otherwise
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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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return isinstance(revision_id, basestring) and revision_id.endswith(':')
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def check_not_reserved_id(revision_id):
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"""Raise ReservedId if the supplied revision_id is reserved"""
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if is_reserved_id(revision_id):
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raise errors.ReservedId(revision_id)
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def ensure_null(revision_id):
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"""Ensure only NULL_REVISION is used to represent the null revision"""
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if revision_id is None:
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symbol_versioning.warn('NULL_REVISION should be used for the null'
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' revision instead of None, as of bzr 0.91.',
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DeprecationWarning, stacklevel=2)
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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
177
a common even when trees are not branches don't represent a single line
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RBC: 20051024: note that when we have two partial histories, this may not
180
be possible. But if we are willing to pretend :)... sure.
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descendants[revision] = {}
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while len(lines) > 0:
190
if line == NULL_REVISION:
195
rev = revision_source.get_revision(line)
196
parents = list(rev.parent_ids)
197
if len(parents) == 0:
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parents = [NULL_REVISION]
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except bzrlib.errors.NoSuchRevision:
203
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)
207
if parent not in descendants:
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descendants[parent] = {}
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descendants[parent][line] = 1
210
if parents is not None:
211
ancestors[line] = set(parents)
214
# The history for revision becomes inaccessible without
215
# actually hitting a no-parents revision. This then
216
# makes these asserts below trigger. So, if root is None
217
# determine the actual root by walking the accessible tree
218
# and then stash NULL_REVISION at the end.
220
descendants[root] = {}
221
# for every revision, check we can access at least
222
# one parent, if we cant, add NULL_REVISION and
224
for rev in ancestors:
225
if len(ancestors[rev]) == 0:
226
raise RuntimeError('unreachable code ?!')
228
for parent in ancestors[rev]:
229
if parent in ancestors:
233
descendants[root][rev] = 1
234
ancestors[rev].add(root)
235
ancestors[root] = set()
236
assert root not in descendants[root]
237
assert root not in ancestors[root]
238
return root, ancestors, descendants
241
def combined_graph(revision_a, revision_b, revision_source):
242
"""Produce a combined ancestry graph.
243
Return graph root, ancestors map, descendants map, set of common nodes"""
244
root, ancestors, descendants = revision_graph(revision_a, revision_source)
245
root_b, ancestors_b, descendants_b = revision_graph(revision_b,
248
raise bzrlib.errors.NoCommonRoot(revision_a, revision_b)
250
for node, node_anc in ancestors_b.iteritems():
251
if node in ancestors:
254
ancestors[node] = set()
255
ancestors[node].update(node_anc)
256
for node, node_dec in descendants_b.iteritems():
257
if node not in descendants:
258
descendants[node] = {}
259
descendants[node].update(node_dec)
260
return root, ancestors, descendants, common
263
def common_ancestor(revision_a, revision_b, revision_source):
265
root, ancestors, descendants, common = \
266
combined_graph(revision_a, revision_b, revision_source)
267
except bzrlib.errors.NoCommonRoot:
268
raise bzrlib.errors.NoCommonAncestor(revision_a, revision_b)
270
distances = node_distances (descendants, ancestors, root)
271
farthest = select_farthest(distances, common)
272
if farthest is None or farthest == NULL_REVISION:
273
raise bzrlib.errors.NoCommonAncestor(revision_a, revision_b)
277
class MultipleRevisionSources(object):
278
"""Proxy that looks in multiple branches for revisions."""
279
def __init__(self, *args):
280
object.__init__(self)
281
assert len(args) != 0
282
self._revision_sources = args
284
def get_revision(self, revision_id):
285
for source in self._revision_sources:
287
return source.get_revision(revision_id)
288
except bzrlib.errors.NoSuchRevision, e:
292
def get_intervening_revisions(ancestor_id, rev_id, rev_source,
293
revision_history=None):
294
"""Find the longest line of descent from maybe_ancestor to revision.
295
Revision history is followed where possible.
297
If ancestor_id == rev_id, list will be empty.
298
Otherwise, rev_id will be the last entry. ancestor_id will never appear.
299
If ancestor_id is not an ancestor, NotAncestor will be thrown
301
root, ancestors, descendants = revision_graph(rev_id, rev_source)
302
if len(descendants) == 0:
303
raise NoSuchRevision(rev_source, rev_id)
304
if ancestor_id not in descendants:
305
rev_source.get_revision(ancestor_id)
306
raise bzrlib.errors.NotAncestor(rev_id, ancestor_id)
307
root_descendants = all_descendants(descendants, ancestor_id)
308
root_descendants.add(ancestor_id)
309
if rev_id not in root_descendants:
310
raise bzrlib.errors.NotAncestor(rev_id, ancestor_id)
311
distances = node_distances(descendants, ancestors, ancestor_id,
312
root_descendants=root_descendants)
314
def best_ancestor(rev_id):
316
for anc_id in ancestors[rev_id]:
318
distance = distances[anc_id]
321
if revision_history is not None and anc_id in revision_history:
323
elif best is None or distance > best[1]:
324
best = (anc_id, distance)
329
while next != ancestor_id:
331
next = best_ancestor(next)
192
def is_null(revision_id):
193
if revision_id is None:
194
symbol_versioning.warn('NULL_REVISION should be used for the null'
195
' revision instead of None, as of bzr 0.90.',
196
DeprecationWarning, stacklevel=2)
197
return revision_id in (None, NULL_REVISION)