Re: Interval aggregate regression failure (expected seems

Have we made any progress on this?

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Tom Lane wrote:
> Michael Glaesemann <grzm(at)seespotcode(dot)net> writes:
> > ... I think this just confirms that there is some kind of rounding (or
> > lack of) in interval_div. Kind of frustrating that it's not visible
> > in the result.
>
> I think the fundamental problem is that the float8 results of division
> are inaccurate, and yet we're assuming that we can (for instance) coerce
> them to integer and get exactly the right answer. For instance, in the
> '41 months'/10 example, I get month_remainder_days being computed as
>
> (gdb) p month_remainder
> $19 = 0.099999999999999645
> (gdb) s
> 2575 result->day += (int32) month_remainder_days;
> (gdb) p month_remainder_days
> $20 = 2.9999999999999893
>
> The only way we can really fix this is to be willing to round off
> the numbers, and I think the only principled way to do that is to
> settle on a specific target accuracy, probably 1 microsecond.
> Then the thing to do would be to scale up all the intermediate
> float results to microseconds and apply rint(). Something like
> (untested)
>
> month_remainder = rint(span->month * USECS_PER_MONTH / factor);
> day_remainder = rint(span->day * USECS_PER_DAY / factor);
> result->month = (int32) (month_remainder / USECS_PER_MONTH);
> result->day = (int32) (day_remainder / USECS_PER_DAY);
> month_remainder -= result->month * USECS_PER_MONTH;
> day_remainder -= result->day * USECS_PER_DAY;
>
> /*
> * Handle any fractional parts the same way as in interval_mul.
> */
>
> /* fractional months full days into days */
> month_remainder_days = month_remainder * DAYS_PER_MONTH;
> extra_days = (int32) (month_remainder_days / USECS_PER_DAY);
> result->day += extra_days;
> /* fractional months partial days into time */
> day_remainder += month_remainder_days - extra_days * USECS_PER_DAY;
>
> #ifdef HAVE_INT64_TIMESTAMP
> result->time = rint(span->time / factor + day_remainder);
> #else
> result->time = rint(span->time * 1.0e6 / factor + day_remainder) / 1.0e6;
> #endif
>
> This might need a few more rint() calls --- I'm assuming that float ops
> with exact integral inputs will be OK, which is an assumption used
> pretty widely in the datetime code, but ...
>
> Per the comment, if we do this here we probably want to make
> interval_mul work similarly.
>
> regards, tom lane

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Bruce Momjian bruce(at)momjian(dot)us
EnterpriseDB http://www.enterprisedb.com

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