You are correct that the pattern .*?foo can be written as (?:[^f]+|f(?!oo))*foo, or the variant you showed. You are correct in assuming that this general principle can be used for practically any regex.

However, you should "unroll the loop" of that regex. Look at these results: 

use Benchmark 'cmpthese';
my $str = "this is the best end";
cmpthese(-5, {
  extra => sub { $str =~ /(?:[^e]+|e(?!nd))*end/ },
  plain => sub { $str =~ /.*?end/ },
});
__END__
     extra:  12019.37/s (n=60818)
     plain:  49615.59/s (n=283305)

         Rate extra plain
extra 12019/s    --  -76%
plain 49616/s  313%    --

[download]
If we unroll the loop, we get better results. 
use Benchmark 'cmpthese';
my $str = "this is the best end";
cmpthese(-5, {
  extra => sub { $str =~ /(?:[^e]+|e(?!nd))*end/ },
  plain => sub { $str =~ /.*?end/ },
  uroll => sub { $str =~ /[^e]*(?:e(?!nd)[^e]*)*end/ },
});
__END__
     extra:  11711.62/s (n=61486)
     plain:  49329.33/s (n=250593)
     uroll:  17985.61/s (n=94964)

         Rate extra uroll plain
extra 11712/s    --  -35%  -76%
uroll 17986/s   54%    --  -64%
plain 49329/s  321%  174%    --

[download]
But the point is, the internal optimization uses less regex op-codes, and so it is faster -- it has the engine do the work itself, it doesn't make the regex do the work.

_____________________________________________________
Jeff[japhy]Pinyan: Perl, regex, and perl hacker, who'd like a (from-home) job
s++=END;++y(;-P)}y js++=;shajsj<++y(p-q)}?print:??;

In reply to Re: Avoiding regex backtracking by japhy
in thread Avoiding regex backtracking by Aristotle

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