Avi Kivity <[EMAIL PROTECTED]> writes: | On Wed, 2005-07-06 at 15:54 +0300, Michael Veksler wrote: | | > > most architectures have different bit representations for +0.0 and -0.0, | > > yet the two values compare equal. | > > | > | > Yet, their sign bit is observable through things like | > assert(a == 0.0); | > assert(b == 0.0); | > 1/(1/a+ 1/b) | > which would give either NaN or 0 depending on the sign | > of a and b. | > | > So do you want one or two copies in the set? | > | what matters is whether the sign bit is observable through the equality | predicate. in the case of the operator==(double, double), it is not | observable, so there should be only one copy in a set.
Yes. That logical framework has some "problems" though. Assume x is NaN, then you would end up with as many xs as you insert in the set. -- Gaby
