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Subject: Re: [RFC] Spaceship operator RFC
From: smalyshev@gmail.com (Stanislav Malyshev)

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