small cleanup patch for e_pow.c
Steve Kargl
sgk at troutmask.apl.washington.edu
Sun May 10 19:01:17 UTC 2015
On Sun, May 10, 2015 at 08:16:14PM +1000, Bruce Evans wrote:
> On Sat, 9 May 2015, Steve Kargl wrote:
> >
> > I don't see how this can be an optimization. The code has the form
> >
> > if (|y| > 2**31) {
> > if (|y| > 2**64) {
> > if (a) return
> > if (b) return
> > }
> > if (a') return
> > if (b') return
> > ...
> > }
> >
> > The difference between a and a' is <= instead of <, and similar for
> > b and b' with >= and >. If either a or b would have return, so will
> > a' and b'. If neither a nor b return, then neither a' nor b' will
> > return. The a' and b' lines were added by das in r141296, where the
> > commit message says the change is for fdlibm 5.3 compatibility.
>
> Hmm, I read the comment and not the code. According to the comment,
> there are no tests a and b. But these are apparently needed.
> Obviously, we must return 1 when x is exactly 1, and it is apparently
> possible for overflow/underflow to occur for values near 1 even when
> y is huge.
>
> > I'll also note that block like 'if (|y|>2**64) { }' is not present
> > in e_powf.c
>
> Apparently just another translation error.
>
> The tests for the double precision case are probably fuzzy just for
> efficiency -- we want to look at only the high word. In float
> precision, we can look at all the bits. Since even 1 bit away from 1
> is relatively large in float precision, overflow/underflow is more
> likely to happen automatically. Indeed, (1+2**-23)**(2**30) is a large
> multiple of FLT_MAX although (1+2**-23)**(2**29) is a small fraction
> of FLT_MAX; N >= 2**30 and above also gives underflow for (1-2**-24)**N;
> N >= 2**62 gives overflow in double precision for (1+2**-52)**N, ...
>
> So 1 is the only numbers near 1 that doesn't give overflow.
>
Thanks for the explanation! That help dislodge a mental block.
To find the magic numbers, it seems I need to consider
(1-2**(-p))**(2**N) = 2**(emin-p) for underflow
(1+2**(-p))**(2**N) = (1-2**(-p))*2**emax for overflow
With p = [24, 53, 64, 113], emin = [-125, -1021, -16381, -16381],
emax = [128, 1024, 16384, 16384], and the use of log(1+z) = z for
|z| << 1, I find
underflow: N = [30.7, 62.5, 77.5, 126.5]
overflow: N = [30.5, 62.5, 77.5, 126.5]
PS: Yes, I'm slowly poking away at powl(). I have the 19 special cases
working for ld80. Now, comes the hard part.
--
Steve
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