Re: Use of C99 extra long double math functions after r236148

On Wed, Jul 25, 2012 at 07:33:07PM +0200, Rainer Hurling wrote:
> On 25.07.2012 19:00 (UTC+2), Steve Kargl wrote:
> >
> >If you actually want to test expl() to see if it is producing
> >a decent result, you need a reference solution that contains
> >a higher precision.  I use mpfr with 256 bits of precision.
> >
> >troutmask:fvwm:kargl[213] ./testl -V 2
> >ULP = 0.3863
> >   x = 2.000000000000000000e+00
> >libm: 7.389056098930650227e+00 0x1.d8e64b8d4ddadcc4p+2
> >mpfr: 7.389056098930650227e+00 0x1.d8e64b8d4ddadcc4p+2
> >mpfr: 7.3890560989306502272304274605750078131803155705518\
> >           47324087127822522573796079054e+00
> >mpfr: 
> >0x7.63992e35376b730ce8ee881ada2aeea11eb9ebd93c887eb59ed77977d109f148p+0
> >
> >The 1st 'mpfr:' line is produced after converting the results
> >fof mpfr_exp() to long double.  The 2nd 'mpfr:' line is
> >produced by mpfr_printf() where the number of printed
> >digits depends on the 256-bit precision.  The last 'mpfr:'
> >line is mpfr_printf()'s hex formatting.  Unfortunately, it
> >does not normalize the hex representation to start with
> >'0x1.', which makes comparison somewhat difficult.
> >
> 
> Many thanks also for this mpfr example. This helps me to understand a 
> little bit more what is going here. So on amd64 even the expl() result 
> is far from what mpfr provides.

Of course!.  MPFR is a multiple precision library.  One specifies
the precision, and mpfr returns a value with that precision.

#include <mpfr.h>

int
main(void)
{
	int i, j[5] = {24, 53, 64, 113, 256};
	mpfr_t x, f;

	for (i = 0; i < 5; i++) {
        /* Set working precision to j[i]. */
		mpfr_inits2(j[i], x, f, MPFR_RNDN); 
		mpfr_set_ui(x, 2, MPFR_RNDN);
		mpfr_exp(f, x,  MPFR_RNDN);
		mpfr_printf("exp(%Re) = %Re\n", x, f);
		mpfr_clears(x, f, NULL);
	}

}

troutmask:fvwm:kargl[222] cc -o z -I/usr/local/include a.c -L/usr/local/lib\
   -lmpfr -lgmp
troutmask:fvwm:kargl[223] ./z
exp(2e+00) = 7.38905621e+00
exp(2e+00) = 7.3890560989306504e+00
exp(2e+00) = 7.3890560989306502274e+00
exp(2e+00) = 7.38905609893065022723042746057500802e+00
exp(2e+00) = 7.38905609893065022723042746057500781\
               3180315570551847324087127822522573796079054e+00



-- 
Steve