Re: [RFC] Combined Comparison (Spaceship) Operator

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