On Thursday, 15 August 2013 at 05:12:35 UTC, Jonathan M Davis
wrote:
On Thursday, August 15, 2013 06:01:21 bsd wrote:
Hi all,
There are multiple values for NaN. It's still a bit surprising
to me that
assert(val is double.nan) failed, but as long as val is still
one of the
values for NaN, it shouldn't matter. It wouldn't have been at
all suprising
though if you had two different NaNs which were generated
through arithmetic
which didn't match. In general, you can't rely on the bit
pattern for NaN.
I also found this confusing:
---
void main() {
assert(double.init is double.nan); // passes expected
assert(double.init is float.init); // passes, unexpected
assert(double.init is real.init); // passes, unexpected
assert(double.init is real.nan); // passes, unexpected
assert(double.init is float.nan); // passes, unexpected
}
---
I don't think these should be passing...should they??
To do the comparison, the same size types must be compared, so
the smaller
floating point types are promoted to the larger type in the
expression, and
evidently, promoting the init value of a smaller floating point
value gives you
the same value as the init value of the larger floating point
value.
- Jonathan M Davis
This explains why (double.nan is float.nan) && (double.nan is
real.nan). I understand that now, thanks.
I found std.math.isNaN in just after I posted, I really should
read the docs first :D
For some reason I thought ('var' is double.nan) did this isNaN
check, sort of like (key in AA) ... but 'in' is in and 'is' is,
well, just is.
Thanks for the help.
