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To: Rainer Hurling <rhurlin@gwdg.de>
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Cc: David Schultz <das@freebsd.org>, freebsd-current@freebsd.org,
	Bruce Evans <bde@freebsd.org>,
	Steve Kargl <sgk@troutmask.apl.washington.edu>
Subject: Re: Use of C99 extra long double math functions after r236148
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On 07/25/12 11:29, Rainer Hurling wrote:

> Many thanks to you three for implementing expl() with r238722 and r238724.
>
> I am not a C programmer, but would like to ask if the following example
> is correct and suituable as a minimalistic test of this new C99 function?
>
>
> //-----------------------------------
> #include <stdio.h>
> #include <math.h>
>
> int main(void)
> {
>    double c = 2.0;
>    long double d = 2.0;
>
>    double e = exp(c);
>    long double f = expl(d);
>
>    printf("exp(%f)  is %.*f\n",  c, 90, e);
>    printf("expl(%Lf) is %.*Lf\n", d, 90, f);
>
>    return 0;
> }
> //-----------------------------------
>
>
> Compiled with 'c99 -o math_expl math_expl.c -lm' and running afterwards
> it gives me:
>
> exp(2.000000)  is
> 7.389056098930650406941822438966482877731323242187500000000000000000000000000000000000000000
>
> expl(2.000000) is
> 7.389056098930650227397942675366948606097139418125152587890625000000000000000000000000000000
>

Just as a point of comparison, here is the answer computed using 
Mathematica:

N[Exp[2], 50]
7.3890560989306502272304274605750078131803155705518

As you can see, the expl solution has only a few digits more accuracy 
that exp.