Re: Error with the leap year

Michael G Schwern wrote:
>It does present a problem... how do you reliably say "same date next year"?

"Same date next year" is exactly what the original poster asked DateTime
for, by using the set_year method, and he got the correct answer: "there
isn't one".  Presumably what he actually wanted was "a year later",
which is a different question.  DateTime does support it:

	$dt->add(years=>1);

In the awkward case, it advances to March 1st:

	$ perl -MDateTime -lwe '
		$dt=DateTime->new(
			year=>2012,
			month=>2,
			day=>29,
			hour=>12,
			minute=>0,
			second=>0,
			time_zone=>"UTC",
		);
		$dt->add(years=>1);
		print $dt->iso8601;
	'
	2013-03-01T12:00:00

Of course, it has the downside that year addition isn't associative:
advancing from a February 29th by four years gives a different result
from advancing by one year four times.

>You can't add 365 days because that falls afoul of the leap year problem.

You can ask DateTime for that too:

	$dt->add(days=>365);

It's nicer, in that day addition *is* associative.  Of course, a quarter
of the time (when the 365 days spans a leap day), the 365-days-later
won't be the same-date-next-year.  Is that what you mean by the "leap
year problem"?

-zefram