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.
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