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Postings from March 2013
-----Original Message-----
From: Nicholas Clark
Sent: Monday, March 25, 2013 12:03 AM
> On Sat, Mar 23, 2013 at 09:15:58PM -0700, Sisyphus wrote:
>
> >
> > Perl correctly evaluates 2**1e15 as infinity.
> > But it evaluates 2**1e16 as 0. For exponents greater than 16, 0
> > is also returned. (This shows up in one of the
> > t/op/sprintf2.t tests as a 'divide-by-zero' error.)
>
> The actual maths should be hitting this rather simple bit of pp.c:
>
> #else
> SETn( Perl_pow( left, right) );
> #endif /* HAS_AIX_POWL_NEG_BASE_BUG */
>
>
> so the question is, does it end up somewhere else? Or does it get there,
> but the C runtime is the problem?
>
> If you compile a C program with the same compiler flags as perl was built
> with, do you get the same results? ie
>
> #include <math.h>
> #include <stdio.h>
>
> int
> main (int argc, char **argv) {
> double power = 1e16;
> printf("2**%g = %g\n", power, pow(2, power));
> return 0;
> }
Bugger .... yes, it's the same result - though the right-wrong threshold for
the C program is 1e9 (instead of 1e15).
I have a number of other C compilers on this box which, I think, use the
very same C runtime. They all seem to do the right thing, so I'm thinking
this is a bug in the particular compiler being used.
Anyway ... sorry for the noise ... feel free to close the bug report.
Cheers,
Rob
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