Object Assignment in PHP

Question

First, sorry for the stupid question, but I was reading an article in php.net and I couldn't understand what exactly it says.

<?php
class SimpleClass
{
    // property declaration
    public $var = 'a default value';

    // method declaration
    public function displayVar() {
        echo $this->var;
    }
}
?>



<?php

$instance = new SimpleClass();

$assigned   =  $instance;
$reference  =& $instance;

$instance->var = '$assigned will have this value';

$instance = null; // $instance and $reference become null

var_dump($instance);
var_dump($reference);
var_dump($assigned);
?>

And this outputs this:

NULL
NULL
object(SimpleClass)#1 (1) {
   ["var"]=>
     string(30) "$assigned will have this value"
}

$instance and $reference point to same place, I get this and I understand why we get NULL and NULL for them.

But what about $assigned? It's it also pointing the place where $instance is stored? Why when we use $instance->var it affects $assigned, but when we set $instance to be null, there is no change for $assigned?

I thought that the all three variables point to one place in the memory, but obviously I'm wrong. Could you please explain me what exactly happens and what is $assigned? Thank you very much!

Answer 1

$reference points to the value of $instance, which is itself a reference to an object. So when you change the value contained by $instance, $reference reflects this change.

$assigned, on the other hand, is a copy of the value of $instance, and independently points to the object itself to which $instance refers. So when the value $instance is updated to point to nothing (i.e. null), $assigned is not affected, since it still points to the object, and doesn't care about the contents of $instance.