RE: last day of month (was MIDRANGE-L Digest V2 #1546)

>Date: Thu, 14 Oct 1999 09:16:48 -0700
>From: Jon Erickson <jerickson@800.com>
>
>Correct me if I'm wrong,
> ...
>
>d                 DS
>d Isodate                         d   Inz
>d  Isoyyyy                       4  0 Overlay(Isodate)
>d  Isomm                         2  0 Overlay(Isodate:6)
>
>c                   Eval      Isoyyyy = Pyear
>c                   Eval      Isomm = Pmonth
>Add 1 month via ADDDUR
>c                   AddDur    1:*Months     Isodate
>
>Subtract 1 day via SUBDUR
>c                   SubDur    1:*Days       Isodate

Sure, that will work.  I was discussing Jon's code to solve the general case of
getting the last day of the month containing a given date.

Jon's code can be made to run a bit more quickly using overlays (you can get rid
of the extract by just setting the days part to "01").  Maybe that's a good
idea,
maybe not - it seems to me that using overlays makes the code a bit more
complex.

If you start with numeric year and month as in your example, your way is as good
as any.

Hmm, I just thought of a way to use numeric year and month input avoiding
overlays, with a single adddur:

D date            s               d       datfmt(*iso) inz(D'0001-01-31')
 * Find the number of months after 0001-01 for the given year and month
 * E.g. for 0002-03, there are 12 + 1 months
C                   eval      Pmonth = ((Pyear - 1) * 12) + (Pmonth - 1)
 * Add them to Jan 31, 0001.  Month-end adjustment will fix up any results
 * where the number of days in the month is less than 31.
C                   adddur    Pmonth:*m     date

With date operations, the contest should be regarding the number of date
opcodes,
not the number of statements.  (Date opcodes are very expensive.)

Barbara Morris


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