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Subject: Re: [RFC] Spaceship operator RFC
From: cmbecker69@gmx.de (Christoph Becker)

Hi!

Stanislav Malyshev wrote:
>
>> 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?

There is the following example in the RFC:

    // only values are compared
    $a = (object) ["a" => "b"];
    $b = (object) ["b" => "b"];
    echo $a <=> $b; // 0

But the actual result is currently 1, not 0.

>> 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.

When there is no "right" value, why not raising E_NOTICE at least.  It
appears to me that returning 1 (as "undefined") without further notice
is too misleading, as it suggests that $a is greater than $b, but that
is not true: ($a > $b === false).

-- 
Christoph M. Becker