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From: John Maddock (john_at_[hidden])
Date: 2005-12-03 06:01:45

> So go figure :-) Optimisation improves the compliance.... bizarre,
> anyway there are a whole heap of Optimisation options that probably
> change this behaviour. There is an entire man page for it separate
> from the main compiler one so I'll look through that to see which ones
> are important. I belive you can't actually tell at compile time if
> denormalised are supported like the code is doing I think you need:
> if (std::numeric_limits<T>::has_denorm == std::denorm_present )
> But thats on a brief inspection, may also want to test using
> denorm_min.

Thanks for the results, looks like I should check that the value isn't zero
before assuming it's a denorm. And yes, I will test denorm_min as well.


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