floating point verification using is?

Stewart Gordon smjg_1998 at yahoo.com
Sun Dec 27 13:37:53 PST 2009

Steven Schveighoffer wrote:
> Are you still working on this?  :)  I think this proves my point.  The 
> compiler does not provide an easy way to compare floats bitwise, so this 
> means convoluted hard-to-write code.  When we have two operators that do 
> equality -- and one of those means bitwise compare in all other contexts 
> -- I think this is a no-brainer.

I entirely agree.

I've identified these cases in which float equality disagrees with 
bitwise equality:
- one is +0, one is -0
- both are NaNs, identically signed
- both are infinity, identically signed

In each case, is just does the same as ==.

Indeed, isIdentical is an ugly workaround for this, apparently created 
instead of sanitising the definition of is to avoid the minuscule amount 
of code breakage that the latter would effect.


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