Re: min() and NaN

If a compare with NaN is always false, how about rewriting it as:
result = ((arg1 < arg2) ? arg2 : arg1).

Or better yet, swap arg1 and arg2 when calling float8smaller.
Use flaost8smaller( current_min, value).

JLL

Tom Lane wrote:
>
> "Michael S. Tibbetts" <mtibbetts(at)head-cfa(dot)cfa(dot)harvard(dot)edu> writes:
> > I'd expect the aggregate function min() to return the minimum, valid
> > numeric value. Instead, it seems to return the minimum value from the
> > subset of rows following the 'NaN'.
>
> Not real surprising given than min() is implemented with float8smaller,
> which does this:
>
> result = ((arg1 > arg2) ? arg1 : arg2);
>
> In most C implementations, any comparison involving a NaN will return
> "false". So when we hit the NaN, we have arg1 = min so far, arg2 = NaN,
> comparison yields false, result is NaN. On the next row, we have
> arg1 = NaN, arg2 = next value, comparison yields false, result is next
> value; and away it goes.
>
> We could probably make it work the way you want with explicit tests for
> NaN in float8smaller, arranged to make sure that the result is not NaN
> unless both inputs are NaN. But I'm not entirely convinced that we
> should make it work like that. The other float8 comparison operators
> are designed to treat NaN as larger than every other float8 value (so
> that it has a well-defined position when sorting), and I'm inclined to
> think that float8smaller and float8larger probably should behave
> likewise. (That actually is the same as what you want for MIN(), but
> not for MAX() ...)
>
> Comments anyone?
>
> regards, tom lane
>
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