On 04/23/2013 07:43 AM, qznc wrote:
> I want to generate a random "double" value, excluding wierdos like NaN
> and Infinity. However, std.random.uniform seems to be useless. I tried
> things like
>
> std.random.uniform( double.min, double.max);
> std.random.uniform(-double.max, double.max);
> std.random.uniform(0.0, double.max);
Since double.max is a valid double value, you may want to call uniform
with a closed range:
uniform!"[]"(-double.max, double.max)
However, that still produces double.inf. :)
> Should that be considered a bug?
I would say yes, it is a bug.
> Nevertheless, any ideas how to work around that issue?
Floating point numbers have this interesting property where the number
of representable values in the range [double.min_normal, 1) is equal to
the number of representable values in the range [1, double.max]. The
number line on this article should help with what I am trying to say:
http://dlang.org/d-floating-point.html
So, to workaround this problem I would suggest simply using the
following range:
uniform(-1.0, 1.0);
One benefit is, now the range is open ended: You don't need to provide
!"[)" to leave 1 out, because it is the default. On the other hand, the
width of the range is now 2.0 so you must keep that in mind when you
scale the value:
Additionally, note that the random numbers in the range between
[-double.min_normal, double.min_normal] and outside of those may have a
different distribution. Test before using. :) (I have no experience with
floating point random numbers.)
Ali
