bisect: avoid copying ancestor list for non-merge commits
During a bisection, hg needs to compute a list of all ancestors for every
candidate commit. This is accomplished via a bottom-up traversal of the set of
candidates, during which each revision's ancestor list is populated using the
ancestor list of its parent(s). Previously, this involved copying the entire
list, which could be very long in if the bisection range was large.
To help improve this, we can observe that each candidate commit is visited
exactly once, at which point its ancestor list is copied into its children's
lists and then dropped. In the case of non-merge commits, a commit's ancestor
list consists exactly of its parent's list plus itself. This means that we can
trivially reuse the parent's existing list for one of its non-merge children,
which avoids copying entirely if that commit is the parent's only child. This
makes bisections over linear ranges of commits much faster.
During some informal testing in the large publicly-available `mozilla-central`
repository, this noticeably sped up bisections over large ranges of history:
Setup:
$ cd mozilla-central
$ hg bisect --reset
$ hg bisect --good 0
$ hg log -r tip -T '{rev}\n'
628417
Test:
$ time hg bisect --bad tip --noupdate
Before:
real 3m35.927s
user 3m35.553s
sys 0m0.319s
After:
real 1m41.142s
user 1m40.810s
sys 0m0.285s
import unittest
try:
from mercurial import rustext
rustext.__name__ # trigger immediate actual import
except ImportError:
rustext = None
else:
from mercurial.rustext import revlog
# this would fail already without appropriate ancestor.__package__
from mercurial.rustext.ancestor import LazyAncestors
from mercurial.testing import revlog as revlogtesting
@unittest.skipIf(
rustext is None,
"rustext module revlog relies on is not available",
)
class RustRevlogIndexTest(revlogtesting.RevlogBasedTestBase):
def test_heads(self):
idx = self.parseindex()
rustidx = revlog.MixedIndex(idx)
self.assertEqual(rustidx.headrevs(), idx.headrevs())
def test_get_cindex(self):
# drop me once we no longer need the method for shortest node
idx = self.parseindex()
rustidx = revlog.MixedIndex(idx)
cidx = rustidx.get_cindex()
self.assertTrue(idx is cidx)
def test_len(self):
idx = self.parseindex()
rustidx = revlog.MixedIndex(idx)
self.assertEqual(len(rustidx), len(idx))
def test_ancestors(self):
idx = self.parseindex()
rustidx = revlog.MixedIndex(idx)
lazy = LazyAncestors(rustidx, [3], 0, True)
# we have two more references to the index:
# - in its inner iterator for __contains__ and __bool__
# - in the LazyAncestors instance itself (to spawn new iterators)
self.assertTrue(2 in lazy)
self.assertTrue(bool(lazy))
self.assertEqual(list(lazy), [3, 2, 1, 0])
# a second time to validate that we spawn new iterators
self.assertEqual(list(lazy), [3, 2, 1, 0])
# let's check bool for an empty one
self.assertFalse(LazyAncestors(idx, [0], 0, False))
if __name__ == '__main__':
import silenttestrunner
silenttestrunner.main(__name__)