revset: stop serializing node when using "%ln"
Turning hundred of thousand of node from node to hex and back can be slow… what
about we stop doing it?
In many case were we are using node id we should be using revision id. However
this is not a good reason to have a stupidly slow implementation of "%ln".
This caught my attention again because the phase discovery during push make an
extensive use of "%ln" or huge set. In absolute, that phase discovery probably
should use "%ld" and need to improves its algorithmic complexity, but improving
"%ln" seems simple and long overdue. This greatly speeds up `hg push` on
repository with many drafts.
Here are some relevant poulpe benchmarks:
### data-env-vars.name = mozilla-try-2023-03-22-zstd-sparse-revlog
# benchmark.name = hg.command.push
# bin-env-vars.hg.flavor = default
# bin-env-vars.hg.py-re2-module = default
# benchmark.variants.explicit-rev = all-out-heads
# benchmark.variants.
issue6528 = disabled
# benchmark.variants.protocol = ssh
# benchmark.variants.reuse-external-delta-parent = default
## benchmark.variants.revs = any-1-extra-rev
before: 44.235070
after: 20.416329 (-53.85%, -23.82)
## benchmark.variants.revs = any-100-extra-rev
before: 49.234697
after: 26.519829 (-46.14%, -22.71)
### benchmark.name = hg.command.bundle
# bin-env-vars.hg.flavor = default
# bin-env-vars.hg.py-re2-module = default
# benchmark.variants.revs = all
# benchmark.variants.type = none-streamv2
## data-env-vars.name = heptapod-public-2024-03-25-zstd-sparse-revlog
before: 10.138396
after: 7.750458 (-23.55%, -2.39)
## data-env-vars.name = mercurial-public-2024-03-22-zstd-sparse-revlog
before: 1.263859
after: 0.700229 (-44.60%, -0.56)
## data-env-vars.name = mozilla-try-2023-03-22-zstd-sparse-revlog
before: 399.484481
after: 346.5089 (-13.26%, -52.98)
## data-env-vars.name = pypy-2024-03-22-zstd-sparse-revlog
before: 4.540080
after: 3.401700 (-25.07%, -1.14)
## data-env-vars.name = tryton-public-2024-03-22-zstd-sparse-revlog
before: 2.975765
after: 1.870798 (-37.13%, -1.10)
# bdiff.py - Python implementation of bdiff.c
#
# Copyright 2009 Olivia Mackall <olivia@selenic.com> and others
#
# This software may be used and distributed according to the terms of the
# GNU General Public License version 2 or any later version.
import difflib
import re
import struct
from typing import (
List,
Tuple,
)
def splitnewlines(text: bytes) -> List[bytes]:
'''like str.splitlines, but only split on newlines.'''
lines = [l + b'\n' for l in text.split(b'\n')]
if lines:
if lines[-1] == b'\n':
lines.pop()
else:
lines[-1] = lines[-1][:-1]
return lines
def _normalizeblocks(
a: List[bytes], b: List[bytes], blocks
) -> List[Tuple[int, int, int]]:
prev = None
r = []
for curr in blocks:
if prev is None:
prev = curr
continue
shift = 0
a1, b1, l1 = prev
a1end = a1 + l1
b1end = b1 + l1
a2, b2, l2 = curr
a2end = a2 + l2
b2end = b2 + l2
if a1end == a2:
while (
a1end + shift < a2end and a[a1end + shift] == b[b1end + shift]
):
shift += 1
elif b1end == b2:
while (
b1end + shift < b2end and a[a1end + shift] == b[b1end + shift]
):
shift += 1
r.append((a1, b1, l1 + shift))
prev = a2 + shift, b2 + shift, l2 - shift
if prev is not None:
r.append(prev)
return r
def bdiff(a: bytes, b: bytes) -> bytes:
a = bytes(a).splitlines(True)
b = bytes(b).splitlines(True)
if not a:
s = b"".join(b)
return s and (struct.pack(b">lll", 0, 0, len(s)) + s)
bin = []
p = [0]
for i in a:
p.append(p[-1] + len(i))
d = difflib.SequenceMatcher(None, a, b).get_matching_blocks()
d = _normalizeblocks(a, b, d)
la = 0
lb = 0
for am, bm, size in d:
s = b"".join(b[lb:bm])
if am > la or s:
bin.append(struct.pack(b">lll", p[la], p[am], len(s)) + s)
la = am + size
lb = bm + size
return b"".join(bin)
def blocks(a: bytes, b: bytes) -> List[Tuple[int, int, int, int]]:
an = splitnewlines(a)
bn = splitnewlines(b)
d = difflib.SequenceMatcher(None, an, bn).get_matching_blocks()
d = _normalizeblocks(an, bn, d)
return [(i, i + n, j, j + n) for (i, j, n) in d]
def fixws(text: bytes, allws: bool) -> bytes:
if allws:
text = re.sub(b'[ \t\r]+', b'', text)
else:
text = re.sub(b'[ \t\r]+', b' ', text)
text = text.replace(b' \n', b'\n')
return text