[r6rs-discuss] Re: [Formal] eq?/eqv? misbehave around NaNs

From: Alan Watson <alan>
Date: Fri Nov 24 11:13:19 2006

Per Bothner wrote:
> If inexacts are boxed, then the implementor can easily implement
> eqv? on inexacts by comparing their bits.

I was coming round to this point of view, but then I realized that you
cannot trivially implement eqv? using a bitwise comparison if there are
inexacts with different representations, say singles and doubles. In
such implementations, eqv? on a single and double would have to convert
both values to a common representation and *then* use either bitwise or
floating-point comparison. I admit that in such an implementation (eqv?
+nan.0 +nan.0) would still be true. However, my point is that eqv? in
this implementation no longer means "has the same bit representation".

Regards,

Alan
Received on Thu Nov 23 2006 - 23:56:29 UTC