On Aug 21, 2013, at 9:55 AM, erik quanstrom <quans...@quanstro.net> wrote:
> On Wed Aug 21 12:09:26 EDT 2013, 9f...@hamnavoe.com wrote:
>>> at least in terms of passing floating point test suites
>>> (like python's) the NaN issue doesn't come up
>>
>> Actually it was a test suite that revealed the NaN errors.
>> I wouldn't think it's something anyone needs in normal
>> day-to-day computation, but sometimes boxes must be ticked.
>
> :-) it is hard to imagine how this is useful. it's not like
> ∑{i→∞}-0 is interesting. at least ∏{i→∞}-0 has an alternating
> sign. (so does it converge with no limit?)
>
> the difference i have seen is a situation like
> atan2(-0, x) ≡ -π
> atan2(+0, x) ≡ pi, ∀ x<0.
>
> any ideas on how this is useful?
See comments by Stephen Canon in
http://stackoverflow.com/questions/1565164/what-is-the-rationale-for-all-comparisons-returning-false-for-ieee754-nan-values
Try this:
#include <u.h>
#include <libc.h>
main(){
double a, b;
setfcr(0);
a = 0.0;
b = a/a;
if(a < b) print(" (a < b)");
if(a <= b) print(" (a <= b)");
if(a == b) print(" (a == b)");
if(a != b) print(" (a != b)");
if(a >= b) print(" (a >= b)");
if(a > b) print(" (a > b)");
if(b < a) print(" (b < a)");
if(b <= a) print(" (b <= a)");
if(b == a) print(" (b == a)");
if(b != a) print(" (b != a)");
if(b >= a) print(" (b >= a)");
if(b > a) print(" (b > a)");
if(b != b) print(" (b != b)");
if(b == b) print(" (b == b)");
print("\n");
return 0;
}
It falsely reports b == b when b is NaN.
