Re^2: Why this code run faster?

/g in scalar context would make every second match fail.

sub f { 'network' =~ /^network$/g }
print f()?1:0, "\n" for 1..6;
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1
0
1
0
1
0
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The failing match should be faster than the succeeding match.

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Re^3: Why this code run faster? by gamache (Friar) on Nov 08, 2007 at 17:31 UTC
Good eyes. When I change the test from `if ($phrase2 =~ /^network$/g) {}` to `while ($phrase2 =~ /^network$/g) {}`, the time went from 2.70s to 5.97s. This is in line with what I'd expect -- if 5E6 successful matches and 5E6 failed matches take 2.7s on my CPU, the `while` loop, consisting of 1E7 successes and 1E7 failures, should take about twice as long.	[reply]
Re^3: Why this code run faster? by Gangabass (Vicar) on Nov 09, 2007 at 01:33 UTC
Thank you guys! You are really help me again. And of course special thanks to ikegami.	[reply]