bdiff: rearrange the "better longest match" code
This is primarily to make the code more managable and prepare for later
changes.
More specific assignments might also be slightly faster, even thought it also
might generate a bit more code.
# discovery.py - protocol changeset discovery functions
#
# Copyright 2010 Matt Mackall <mpm@selenic.com>
#
# This software may be used and distributed according to the terms of the
# GNU General Public License version 2 or any later version.
from __future__ import absolute_import
import collections
from .i18n import _
from .node import (
nullid,
short,
)
from . import (
error,
)
def findcommonincoming(repo, remote, heads=None, force=False):
"""Return a tuple (common, fetch, heads) used to identify the common
subset of nodes between repo and remote.
"common" is a list of (at least) the heads of the common subset.
"fetch" is a list of roots of the nodes that would be incoming, to be
supplied to changegroupsubset.
"heads" is either the supplied heads, or else the remote's heads.
"""
knownnode = repo.changelog.hasnode
search = []
fetch = set()
seen = set()
seenbranch = set()
base = set()
if not heads:
heads = remote.heads()
if repo.changelog.tip() == nullid:
base.add(nullid)
if heads != [nullid]:
return [nullid], [nullid], list(heads)
return [nullid], [], heads
# assume we're closer to the tip than the root
# and start by examining the heads
repo.ui.status(_("searching for changes\n"))
unknown = []
for h in heads:
if not knownnode(h):
unknown.append(h)
else:
base.add(h)
if not unknown:
return list(base), [], list(heads)
req = set(unknown)
reqcnt = 0
# search through remote branches
# a 'branch' here is a linear segment of history, with four parts:
# head, root, first parent, second parent
# (a branch always has two parents (or none) by definition)
unknown = collections.deque(remote.branches(unknown))
while unknown:
r = []
while unknown:
n = unknown.popleft()
if n[0] in seen:
continue
repo.ui.debug("examining %s:%s\n"
% (short(n[0]), short(n[1])))
if n[0] == nullid: # found the end of the branch
pass
elif n in seenbranch:
repo.ui.debug("branch already found\n")
continue
elif n[1] and knownnode(n[1]): # do we know the base?
repo.ui.debug("found incomplete branch %s:%s\n"
% (short(n[0]), short(n[1])))
search.append(n[0:2]) # schedule branch range for scanning
seenbranch.add(n)
else:
if n[1] not in seen and n[1] not in fetch:
if knownnode(n[2]) and knownnode(n[3]):
repo.ui.debug("found new changeset %s\n" %
short(n[1]))
fetch.add(n[1]) # earliest unknown
for p in n[2:4]:
if knownnode(p):
base.add(p) # latest known
for p in n[2:4]:
if p not in req and not knownnode(p):
r.append(p)
req.add(p)
seen.add(n[0])
if r:
reqcnt += 1
repo.ui.progress(_('searching'), reqcnt, unit=_('queries'))
repo.ui.debug("request %d: %s\n" %
(reqcnt, " ".join(map(short, r))))
for p in xrange(0, len(r), 10):
for b in remote.branches(r[p:p + 10]):
repo.ui.debug("received %s:%s\n" %
(short(b[0]), short(b[1])))
unknown.append(b)
# do binary search on the branches we found
while search:
newsearch = []
reqcnt += 1
repo.ui.progress(_('searching'), reqcnt, unit=_('queries'))
for n, l in zip(search, remote.between(search)):
l.append(n[1])
p = n[0]
f = 1
for i in l:
repo.ui.debug("narrowing %d:%d %s\n" % (f, len(l), short(i)))
if knownnode(i):
if f <= 2:
repo.ui.debug("found new branch changeset %s\n" %
short(p))
fetch.add(p)
base.add(i)
else:
repo.ui.debug("narrowed branch search to %s:%s\n"
% (short(p), short(i)))
newsearch.append((p, i))
break
p, f = i, f * 2
search = newsearch
# sanity check our fetch list
for f in fetch:
if knownnode(f):
raise error.RepoError(_("already have changeset ")
+ short(f[:4]))
base = list(base)
if base == [nullid]:
if force:
repo.ui.warn(_("warning: repository is unrelated\n"))
else:
raise error.Abort(_("repository is unrelated"))
repo.ui.debug("found new changesets starting at " +
" ".join([short(f) for f in fetch]) + "\n")
repo.ui.progress(_('searching'), None)
repo.ui.debug("%d total queries\n" % reqcnt)
return base, list(fetch), heads