NaN doesn't test as == anything, even another NaN.
I was wondering:
unless ($x <= $inf) { print "\$x is definitely a NaN\n" }
else { print "\$x is definitely NOT a NaN\n" }
Is that reliable ? ... assuming, of course, that $inf is in fact + infinity, and that there are no bugs present.
Seems to me that ought to be good enough.
Cheers,
Rob
| [reply]
[d/l] |
$isnan = ($x != $x);
$isnan = ! defined($x <=> 0);
| [reply]
[d/l] |
$isnan = ! defined($x <=> 0);
Ooow ... wasn't aware that perl did that.
In the mpfr library, the mpfr_cmp(x,y) function will return -1 if x<y, +1 if x>y. Otherwise it returns 0 (and sets the erange flag iff one/both of the arguments is a nan).
So, I'm left wondering whether, in Math::MPFR (which more-or-less wraps the mpfr library), the spaceship operator (which overloads the mpfr_cmp function) should:
a) adhere to mpfr library behaviour and return 0 and set the erange flag when nans are involved;
or
b) adhere to perl behaviour and return undef when nans are involved.
If b), then there's also the question of whether the erange flag should still be set.
Interested to hear any thoughts on that.
Cheers,
Rob
| [reply] |
