v8/tools/regexp-sequences.py
Patrick Thier 6612943010 [regexp] Bytecode peephole optimization
Bytecodes used by the regular expression interpreter often occur in
specific sequences. The number of dispatches in the interpreter can be
reduced if those sequences are combined into a single bytecode.

This CL adds a peephole optimization pass for regexp bytecodes.
This pass checks the generated bytecode for pre-defined sequences that
can be merged into a single bytecode.

With the currently implemented bytecode sequences a speedup of 1.12x on
regex-dna and octane-regexp is achieved.

Bug: v8:9330
Change-Id: I827f93273a5848e5963c7e3329daeb898995d151
Reviewed-on: https://chromium-review.googlesource.com/c/v8/v8/+/1813743
Commit-Queue: Patrick Thier <pthier@google.com>
Reviewed-by: Peter Marshall <petermarshall@chromium.org>
Reviewed-by: Jakob Gruber <jgruber@chromium.org>
Cr-Commit-Position: refs/heads/master@{#63992}
2019-09-26 14:57:37 +00:00

68 lines
1.8 KiB
Python
Executable File

#!/usr/bin/env python
# Copyright 2019 the V8 project authors. All rights reserved.
# Use of this source code is governed by a BSD-style license that can be
# found in the LICENSE file.
"""
python %prog trace-file
Parses output generated by v8 with flag --trace-regexp-bytecodes and generates
a list of the most common sequences.
"""
from __future__ import print_function
import sys
import re
import collections
def parse(file, seqlen):
# example:
# pc = 00, sp = 0, curpos = 0, curchar = 0000000a ..., bc = PUSH_BT, 02, 00, 00, 00, e8, 00, 00, 00 .......
rx = re.compile(r'pc = (?P<pc>[0-9a-f]+), sp = (?P<sp>\d+), '
r'curpos = (?P<curpos>\d+), curchar = (?P<char_hex>[0-9a-f]+) '
r'(:?\.|\()(?P<char>\.|\w)(:?\.|\)), bc = (?P<bc>\w+), .*')
total = 0
bc_cnt = [None] * seqlen
for i in xrange(seqlen):
bc_cnt[i] = {}
last = [None] * seqlen
with open(file) as f:
l = f.readline()
while l:
l = l.strip()
if l.startswith("Start bytecode interpreter"):
for i in xrange(seqlen):
last[i] = collections.deque(maxlen=i+1)
match = rx.search(l)
if match:
total += 1
bc = match.group('bc')
for i in xrange(seqlen):
last[i].append(bc)
key = ' --> '.join(last[i])
bc_cnt[i][key] = bc_cnt[i].get(key,0) + 1
l = f.readline()
return bc_cnt, total
def print_most_common(d, seqlen, total):
sorted_d = sorted(d.items(), key=lambda kv: kv[1], reverse=True)
for (k,v) in sorted_d:
if v*100/total < 1.0:
return
print("{}: {} ({} %)".format(k,v,(v*100/total)))
def main(argv):
max_seq = 7
bc_cnt, total = parse(argv[1],max_seq)
for i in xrange(max_seq):
print()
print("Most common of length {}".format(i+1))
print()
print_most_common(bc_cnt[i], i, total)
if __name__ == '__main__':
main(sys.argv)