6612943010
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}
68 lines
1.8 KiB
Python
Executable File
68 lines
1.8 KiB
Python
Executable File
#!/usr/bin/env python
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# Copyright 2019 the V8 project authors. All rights reserved.
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# Use of this source code is governed by a BSD-style license that can be
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# found in the LICENSE file.
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"""
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python %prog trace-file
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Parses output generated by v8 with flag --trace-regexp-bytecodes and generates
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a list of the most common sequences.
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"""
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from __future__ import print_function
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import sys
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import re
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import collections
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def parse(file, seqlen):
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# example:
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# pc = 00, sp = 0, curpos = 0, curchar = 0000000a ..., bc = PUSH_BT, 02, 00, 00, 00, e8, 00, 00, 00 .......
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rx = re.compile(r'pc = (?P<pc>[0-9a-f]+), sp = (?P<sp>\d+), '
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r'curpos = (?P<curpos>\d+), curchar = (?P<char_hex>[0-9a-f]+) '
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r'(:?\.|\()(?P<char>\.|\w)(:?\.|\)), bc = (?P<bc>\w+), .*')
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total = 0
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bc_cnt = [None] * seqlen
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for i in xrange(seqlen):
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bc_cnt[i] = {}
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last = [None] * seqlen
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with open(file) as f:
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l = f.readline()
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while l:
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l = l.strip()
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if l.startswith("Start bytecode interpreter"):
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for i in xrange(seqlen):
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last[i] = collections.deque(maxlen=i+1)
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match = rx.search(l)
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if match:
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total += 1
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bc = match.group('bc')
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for i in xrange(seqlen):
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last[i].append(bc)
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key = ' --> '.join(last[i])
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bc_cnt[i][key] = bc_cnt[i].get(key,0) + 1
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l = f.readline()
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return bc_cnt, total
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def print_most_common(d, seqlen, total):
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sorted_d = sorted(d.items(), key=lambda kv: kv[1], reverse=True)
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for (k,v) in sorted_d:
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if v*100/total < 1.0:
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return
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print("{}: {} ({} %)".format(k,v,(v*100/total)))
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def main(argv):
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max_seq = 7
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bc_cnt, total = parse(argv[1],max_seq)
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for i in xrange(max_seq):
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print()
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print("Most common of length {}".format(i+1))
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print()
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print_most_common(bc_cnt[i], i, total)
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if __name__ == '__main__':
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main(sys.argv)
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