OOP 4.0 -> 5.0 reference

ReSpawN · 04-29-2008, 05:01 PM

Sup.

Hmm, I was scrolling through a very nice document about GPS PHP tracking and tracing and we came across two very stange variables. An instance defined and a variable in a function or class.

PHP Code:


			
class mySQL {
    function foobar ( $foo, &$bar) {
       $this->foo = $foo;
        $this->bar = &$bar;
    }
}

Anyways, I think you get my drift. Same goes for a class that directly calls to a function, correct me if I am wrong but this is how I, sort of, remember it.
echo instanceName::&variable->foo;

So, could somebody explain those combinations? Same goes for:

PHP Code:


			
$foo = $bar;
$wtf =& $bar;

And a little example of the above:

PHP Code:


			
$a = new A();
$a->foo();
A::foo();
$b = new B();
$b->bar();
B::bar();

Source: PHP: The Basics - Manual

delayedinsanity · 04-29-2008, 05:26 PM

Hmm, not sure if I followed you there, but if I did, I think you're asking about a) references (the & signed being used there), and b) Paamayim Nekudotayim, aka the scope resolution operator used to access static methods of a class, or to (as is allowed for some reason) simply access a public method of that class without creating a new instance of said class.

PHP Code:


			
class staticReference

{

    static $var;



    static function setVar ($var) {

        self::$var = $var;

    }

} // staticReference



staticReference::setVar("dude!");

$reference =& staticReference::$var;



echo staticReference::$var;

echo $reference;



staticReference::setVar("totally.");

echo $reference;



$reference = "whoah.";

echo staticReference::$var;

If I gathered what you were asking correctly, that should be a good demonstration of what's going on. If not, I probably look pretty dumb right now.

staticReference has two parts, a static variable and a method that you can use to set that variable. Both are declared static, so are meant to be used with the scope resolution operator (::), without having to call an instance of the class into existence. Notice how the methods are all being called, but a new statement has never been made?

As for the reference, it creates a pointer of sorts to the original data. If you make a reference to staticReference::$var, whenever you change staticReference::$var, so does the reference you created since it's pointing to the original data, and not a copy of it. Same goes visa versa, it's all accessing the same data.

Did that help at all? It's early here.
-m

ReSpawN · 04-30-2008, 08:02 PM

It's a lot to take in, considering that's all new to me (OOP5) but I think I'll manage. If not, the questions will be asked here of course.