RE: [PHP-DEV] 5.4's New De-referencing plus assignment

IT would probably be even more convenient to just say if( (new Validator())->isValid($value) ){ } No reason to just temporaraly assign $validator. Dmitri Snytkine Web Developer Ultra Logistics, Inc. Phone: (888) 220-4640 x 2097 Fax: (888) 795-6642 E-Mail: dsnytkine@ultralogistics.com Web: www.ultralogistics.com "A Top 100 Logistics I.T. Provider in 2011" -----Original Message----- From: Matthew Weier O'Phinney [mailto:weierophinney@php.net] Sent: Friday, December 02, 2011 11:50 AM To: internals@lists.php.net Subject: Re: [PHP-DEV] 5.4's New De-referencing plus assignment On 2011-12-01, Anthony Ferrara <ircmaxell@gmail.com> wrote: > On Thu, Dec 1, 2011 at 12:34 PM, Ralph Schindler ><ralph@ralphschindler.com> wrote: <snip> > > needs to somehow guarantee that all methods of the type $foo will return > > $this.  (BTW, this is not an argument for my feature as much as its an > > argument as much as its one for "if we're going to do something, why not do > > it correctly in the first place".)  The correct path here, IMO, would be to > > simply carry the expression result (since we're using '(' expr > > ')' out and > > allow dereferencing on whatever comes out of it. > > > > I would argue though that your syntax is completely possible today: > > $foo = new Foo; > $foo->bar(); > > What's the reason to put that in a single line? Aside from terseness, > is there any other benefit? With the new dereference, one benefit is > that no variable is populated when none is needed. But in your case, > you need both variables... Here's another example. We have validator classes. Typically, you call isValid() to see if a value validates. If it does, you have no more use for the validator. If it _doesn't_, however, you'll want to get the error messages. I could see the following as being a nice, succinct way to use validators: if (!(($validator = new SomeValidator())->isValid($value))) { // Validation failed, get messages... $view->assign('errors' => $validator->getMessages()); return $view->render('error'); } // validation passed, do something... Yes, this could be written as follows: $validator = new SomeValidator(); if (!$validator->isValid($value)) { // ... } // ... However, I can see some folks not really wanting that variable declaration if they won't be using it outside the conditional. -- Matthew Weier O'Phinney Project Lead | matthew@zend.com Zend Framework | http://framework.zend.com/ PGP key: http://framework.zend.com/zf-matthew-pgp-key.asc -- PHP Internals - PHP Runtime Development Mailing List To unsubscribe, visit: http://www.php.net/unsub.php