std::isfinite broken?

[Paolo Carlini]

Neal Becker wrote:
> I found that compiling without -ffast-math would allow std::isfinite to
> work.  Sorry if the statement was confusing.  The code looks something
> like:
>
> [calculate x]
> if (not isfinite (x))
>   throw std::runtime_error ("blah")
>   
Well, -ffast-math implies -ffinite-math-only, I think you have to pass 
-fno-finite-math-only if you want to deal correctly with infinities.

Paolo.