Damian Conway wrote: > > But I assume that == means numerically equal (and here I could be > > wrong). If what I assume is true however, then anything which doesn't > > have any numerical meaning, numerically compared to anything (even to > > itself) should not return the misleading result that the two compared > > values are numerically equal. > > > > Then again, if you tell me that == operator doesn't mean "numerically > > equal", I will agree that NaN==NaN should be true even considering that > > 'cat'=='dog' will also be true. > > But 'cat'=='dog' *is* true. Numerically, they *are* equal. > They are equally not numbers. One should certainly get a warning > (and one will if warnings are enabled), but this > expression shouldn't return false.
OK, now I see your point even better. The difference in our points is that I suggested that numerical comparison of NaNs shouldn't make sense. But yes, two NaNs are equally not numbers. Now the question is if being a NaN should be the special case of numerical meaning or the lack of any numerical meaning. I don't think that the one idea is more correct than another, maybe that's the matter of taste, however they are mutually exclusive, so one of them has to be chosen. I joined this thread, because I thought that assuming that NaN!=NaN idea is just ugly and thus not worth deeper discussion, was oversimplifying and the problem deserves a little more of controversy. However I wasn't suspecting that NaN semantics will grow to one of the most important threads here. > Sigh. I *do* see your point of view (Laziness), but I still have immense > difficulty with the notion that: > > $x == NaN > > doesn't return true if $x contains NaN. Well, Laziness is very important to me, but here it won't be hurt much, as in fact $x==$y!=NaN wouldn't be used very often. Only when numerically comparing two unknown values, and I can't remember when I did it last time. Maybe when sorting but with sorting it doesn't matter if every NaN is equal or not. Hmmm, maybe NaN==NaN would make starship operator simpler..? Anyway, even with NaN==NaN 'text'!=0, unlike Perl 5, where 'text'==0, which could've been the main problem with == operator. - RaFaL Pocztarski, [EMAIL PROTECTED]
