Php By Reference

Question

Can someone please explain what the "&" does in the following:

class TEST {

}

$abc =& new TEST();

I know it is by reference. But can someone illustrate why and when I would need such a thing? Or point me to a url where this is explained well. I am unable to grasp the concept.

Thank you very much.

Answer 1

As I understand it, you're not asking about PHP references in general, but about the $foo =& new Bar(); construction idiom.

This is only seen in PHP4 as the usual $foo = new Bar() stores a copy of the object. This generally goes unnoticed unless the class stored a reference to $this in the constructor. When calling a method on the returned object later on, there would be two distinct copies of the object in existence when the intention was probably to have just one.

Consider this code where the constructor stores a reference to $this in a global var

class Bar {
    function Bar(){
       $GLOBALS['copy']=&$this;
        $this->str="hello";
    }

}

//store copy of constructed object
$x=new Bar;
$x->str="goodbye";

echo $copy->str."\n"; //hello
echo $x->str."\n"; //goodbye

//store reference to constructed object
$x=&new Bar;
$x->str="au revoir";

echo $copy->str."\n"; //au revoir
echo $x->str."\n"; //au revoir

In the first example, $x and $copy refer to different instances of Foo, but in the second they are the same.

Answer 2

Firstly, you don't really need to use it if you are using PHP 5, in PHP 5 all objects are passed by reference by default.

Secondly, when you assign an object to a variable name, either by creation, passing in a parameter, or setting a variable value, you are either doing so by reference or value.

Passing by reference means you pass the actual memory reference for the object, so say you passed an object as a parameter to a function, any changes that function makes to that variable will be reflected in the parent method as well, you are actually changing the state of that object in memory.

The alternative, to pass by value means you pass a copy of that object, not the memory reference, so any changes you make, will not be reflected in the original.