Re: [RFC] Spaceship operator RFC

Hi! > A discrepancy between the actual behavior of the spaceship operator and > an example given in its RFC has been reported recently (bug #69466[1]). I'm not sure what the discrepancy is, could you explain? > To me it appears that there is a contradiction in the RFC (see my > comment on the respective report), which would have to be resolved one > way or another. I don't see the contradiction. The objects in your example are not comparable, since $a < $b and $b < $a are both not true, 1 is returned. Returning 0 would be wrong of course since these objects are not equal. In fact, there's no "right" value in this case as objects are not comparable - so the result is undefined. In this case, undefined is 1. -- Stas Malyshev smalyshev@gmail.com