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Date: Thu, 22 Jan 2015 23:52:40 +0000
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To: Larry Garfield <larry@garfieldtech.com>
Subject: Re: [PHP-DEV] [RFC] Combined Comparison (Spaceship) Operator
From: ajf@ajf.me (Andrea Faulds)

Hi again,

> On 22 Jan 2015, at 23:47, Larry Garfield <larry@garfieldtech.com> =
wrote:
>=20
> Assuming it's accurate and I'm understanding you correctly, I think =
the following would be a sufficient statement for the RFC:
>=20
> -----
> The behavior of this operator with mixed types is such that the =
following two snippets produce the exact same result in all cases:
>=20
> return $a <=3D> $b;
>=20
> if ($a > $b) {
>  return -1;
> }
> elseif ($a < $b) {
>  return 1;
> }
> else {
>  return 0;
> }
> -----
>=20
> Or maybe that's implicit in the existing link, I'm not sure.  (Now =
that I read it over again.) Basically, in order to understand the more =
esoteric type juggling that may occur you'd need to mentally expand it =
to its "uncompressed form", for which the rules are already defined (if =
possibly quirky in some cases).

Yes, that=E2=80=99s correct. Although it would be more accurate to say =
that ($a < $b) is just ($a <=3D> $b =3D=3D=3D -1), ($a > $b) is just ($a =
<=3D> $b =3D=3D=3D 1) and similar things for >=3D and <=3D, as that=E2=80=99=
s how the comparison operators work internally (with the exception of =3D=3D=
 which also has a fast case).

I think I=E2=80=99ll add something along those lines to the RFC.

Thanks.
--
Andrea Faulds
http://ajf.me/