Re: Number of bits, length of RegExp

By short, are you really meaning fast, or do you just mean fewest characters? If you want speed, try using non-capturing parenthesis (?:expression) -- it'll save perl some effort and make your search faster. For an even number of ones, I'd actually go with

$foo =~ tr/0//d;
length($foo);
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and handle an odd number similarly. It's not a regex, but it should do the job, and it's very short. If you really want to do a regex, try

/^(?:0*10*1)*/  # Matches evens
/^0*1(?:0*10*1)*/ # Odds
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As you can see, I think your one to match evens is good.

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Re: Re: Number of bits, length of RegExp by diotalevi (Canon) on Apr 21, 2003 at 14:58 UTC
I wouldn't have let tr delete anything - just count the number of ones and then test that number for oddness. So I'd say this as `$is_odd = $foo =~ tr/1// & 1; $is_even = $foo =~ tr/1// ^ 1;`. This only tests bit zero's value and in theory is speedy. I dunno if it actually is.	[reply] [d/l]
Re: Re: Re: Number of bits, length of RegExp by Improv (Pilgrim) on Apr 21, 2003 at 15:02 UTC
Out of academic curiosity, does anyone know if scalar construction is likely to be more or less expensive than scalar modification? As a matter of fact, it would be really interesting to see performance analyses of many common perl operations.. Maybe they're already out there.. Any pointers, anyone?	[reply]