Re: Making Lemonade (Floating-point format changes)

Michael S wrote:

On Tue, 21 May 2024 05:49:55 -0000 (UTC)
Thomas Koenig <tkoenig@netcologne.de> wrote:
 

BGB <cr88192@gmail.com> schrieb:
>

Granted, they are not necessarily the option one would go if they
wanted "cheapest possible FPU that is still good enough to be
usable".
>
Though, the point at which an FPU manages to suck badly enough that
one needs to resort to software emulation to make software work, is
probably a lower limit.
>
>
Luckily, "uses 754 formats, but with aggressive cost cutting" can
be "good enough", and so long as they more-or-less deliver a full
width mantissa, and can exactly compute exact-value calculations,
most software is generally going to work.

>
This will require extensive testing and possibly modification for
a lot of software ported to such a system.  This will drive up
the total cost, presumably far more than any hardware savings.
>

But OTOH, if 1.0+2.0 gives 2.999999, that is, not good enough, so
there is a lower limit here.

>
An example of a more interesting question is
>
   if (a >= 0.) {
     if (b >= 0) {
       if (a + b < a) {
         printf("We should never get here!\n);
abort();
       }
     }
   }

 If I am not mistaken, that should hold on VAX, which has floating-point
very close to BGB ideal. It looks like it would hold even on less
robust formats, like IBM's hex float. I wonder where it is not true?
 The biggest difference between IEEE and VAX is that on IEEE when (a > b)
then (a - b > 0) while on VAX (a - b >= 0).
 Of course, IEEE has non-intuitive cases as well.
if (!(a < 0)) {
   if (!(b < 0)) {
     if (!(a + b >= a)) {
       printf("It's IEEE 754, baby!\n);
     }
   }
}
 

What happens when a and/or b is a NaN?
Comparisons with NaN should return false, so !(NaN < 0) will be true (and the same for b), while !(NaN+b >= NaN) will also return true.
Is that what you were thinking of?
Terje
--
- <Terje.Mathisen at tmsw.no>
"almost all programming can be viewed as an exercise in caching"