Re: Re: function call chaining

Eddie Drapkin wrote: > On Tue, Jan 19, 2010 at 11:05 AM, Stanislav Malyshev <stas@zend.com> wrote: >> The second was next on my list, while the first seems to me kind of exotic - >> why create object only to call one method and immediately drop it? Why this >> method is not static then? > > > Why would this imply "dropping" the object? > > This: > $foo = (new bar())->someSetter(); > Looks a lot better than this > $foo = new bar(); > $foo->someSetter(); The second version is much clearer. You know exactly what $foo is. In the shortened version you have no idea what $foo is without reading the code for the someSetter() method. On first glance I would assume that $foo would be the success/failure return of the setter and that the object is dropped. -Rasmus