Tom Lane wrote:
I've also confirmed that the problem is in interval_div; you can
reproduce the failure with
select '41 years 1 mon 11 days'::interval / 10;
which should give '4 years 1 mon 9 days 26:24:00', but when
timestamp.o is compiled with "-mcpu=pentium4 -march=pentium4",
you get '4 years 1 mon 10 days 02:24:00'. --enable-integer-datetimes
is not relevant because the interesting part is all double/integer
arithmetic.
Looking at this, though, I wonder if the pentium4 answer isn't "more
correct". If I'm doing the math by hand correctly, what we end up
with is having to cascade 3/10ths of a month down into the days field,
and since the conversion factor is supposed to be 30 days/month, that
should be exactly 9 days. Plus the one full day from the 11/10 days
gives 10 days. I think what is happening on all the non-Pentium
platforms is that (3.0/10.0)*30.0 is producing something just a shade
less than 9.0, whereas the Pentium gives 9.0 or a shade over, possibly
due to rearrangement of the calculation. I think we can still file
this
as a compiler bug, because I'm pretty sure the C spec does not allow
rearrangement of floating-point calculations ... but we might want to
think about tweaking the code's roundoff behavior just a bit.
I took a naive look at this, but have been unable to come up with a
satisfactory solution. Here's an even simpler example that exhibits
the behavior:
# select '41 mon'::interval / 10;
?column?
------------------------
4 mons 2 days 24:00:00
And one using a non-integer divisor:
# select '2 mon'::interval / 1.5;
?column?
-----------------------
1 mon 9 days 24:00:00
Here's the relevant code in interval_div (timestamp.c c2573).
month_remainder holds the fractional part of the months field of the
original interval (41) divided by the divisor (10) as a double.
/* fractional months full days into days */
month_remainder_days = month_remainder * DAYS_PER_MONTH;
result->day += (int32) month_remainder_days;
/* fractional months partial days into time */
day_remainder += month_remainder_days - (int32) month_remainder_days;
#ifdef HAVE_INT64_TIMESTAMP
result->time = rint(span->time / factor + day_remainder *
USECS_PER_DAY);
#else
result->time = span->time / factor + day_remainder * SECS_PER_DAY;
#endif
My understanding is that as month_remainder is a float (as is
month_remainder_days), month_remainder_days may be equal to 24 hours
after rounding. As we're converting from months to days, and from
days to time, rather than from months to time directly, we're
assuming that we should only have time less than 24 hours remaining
in the month_remainder_days when it's added to day_remainder. If
month_remainder_days is equal to 24 hours, we should increment result-
>day rather than carrying that down into result->time.
With that in mind, I produced this hack:
#ifdef HAVE_INT64_TIMESTAMP
month_remainder_seconds = month_remainder_day_fraction *
USECS_PER_DAY;
if ( USECS_PER_DAY == rint(month_remainder_seconds) )
result->day++;
else if ( USECS_PER_DAY == - rint(month_remainder_seconds) )
result->day--;
else day_remainder += month_remainder_day_fraction;
result->time = rint(span->time / factor + day_remainder *
USECS_PER_DAY);
#else
month_remainder_seconds = month_remainder_day_fraction *
SECS_PER_DAY;
if ( SECS_PER_DAY == (int32) month_remainder_seconds )
result->day++;
else if ( SECS_PER_DAY == - (int32) month_remainder_seconds)
result->day--;
else day_remainder += month_remainder_day_fraction;
result->time = span->time / factor + day_remainder * SECS_PER_DAY;
#endif
It works okay with --enable-integer-datetimes
postgres=# select '41 mon'::interval/10;
?column?
---------------
4 mons 3 days
(1 row)
postgres=# select '2 mon'::interval/1.5;
?column?
---------------
1 mon 10 days
(1 row)
It also changes the result of the aggregate test for intervals, but I
think that's to be expected.
*** ./expected/interval.out Tue Mar 7 07:49:17 2006
--- ./results/interval.out Thu Jun 22 22:52:41 2006
***************
*** 218,224 ****
select avg(f1) from interval_tbl;
avg
-------------------------------------------------
! @ 4 years 1 mon 9 days 28 hours 18 mins 23 secs
(1 row)
-- test long interval input
--- 218,224 ----
select avg(f1) from interval_tbl;
avg
-------------------------------------------------
! @ 4 years 1 mon 10 days 4 hours 18 mins 23 secs
(1 row)
-- test long interval input
But doesn't work without --enable-integer-datetimes. It still gives
the same answer as before. I don't have a firm grasp of floating
point math so I'm fully aware this might be entirely the wrong way to
go about this. It definitely feels kludgy (especially my sign-
handling), but I submit this in the hope it moves the discussion
forward a little. If someone wanted to point me in the right
direction, I'll try to finish this up, updating the regression test
and adding a few more to test this more thoroughly.
Thanks!
Michael Glaesemann
grzm myrealbox com
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