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Hi Jeff,
> Another astonishment is that the algorithm described makes P1Y1D >
> P365D indeterminate, even though in our naive understanding, even
> including the idea of leap year, a year and a day is always at least
> 366 days.
I realise that I wasn't clear in my earlier mails about this, but I
was only suggesting comparing durations by adding them both to a
particular date and comparing the results when the result of comparing
the durations directly is undefined.
Your example of P1Y1D > P365D isn't undefined, however, as I
understand it. P1Y is between 365 and 366 days. P1Y1D is therefore
between 366 and 367 days. P1Y1D therefore *must* be more than P365D.
On the other hand, P1Y1D > P366D, for example, is undefined if you
just compare the durations, because P1Y1D *could* be more than P366D.
At that point, *one way* of resolving it would be to add them both to
today's date (or another date of your choosing) and comparing the
resulting dates.
Cheers,
Jeni
---
Jeni Tennison
http://www.jenitennison.com/
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