Re: Re: Re: Last day of the month. Any shorter

Still not sure why you'd want to. Just doing the leap-year check by hand is longer than some of the entire solutions presented.

And also:

use Time::Local;
sub is_leapyear {
  1 == (gmtime(timegm(0,0,1,28,1,$_[0])+24*60*60))[4]
}
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(But this has severe limitations on the range of valid dates input.)

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Re: Re: Re: Re: Last day of the month. Any shorter by greenFox (Vicar) on May 20, 2004 at 05:53 UTC
I thought you were indicating there was a specific risk in implementing the usual algorithm and I wanted to know what it was. The gmtime-timegm method you give is a neat trick but I can't see any good reason to prefer it. It really needs code to catch years outside the epoch range, or at a minimum a comment to highlight it's limitation. -- Do not seek to follow in the footsteps of the wise. Seek what they sought. -Basho	[reply]

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Re: Re: Re: Re: Last day of the month. Any shorter
by greenFox (Vicar) on May 20, 2004 at 05:53 UTC

I thought you were indicating there was a specific risk in implementing the usual algorithm and I wanted to know what it was.

The gmtime-timegm method you give is a neat trick but I can't see any good reason to prefer it. It really needs code to catch years outside the epoch range, or at a minimum a comment to highlight it's limitation.

--
Do not seek to follow in the footsteps of the wise. Seek what they sought. -Basho

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