将一个类的实例分配给php中的新对象

Question

Following is an example in manual:

<?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);
 ?>

I cannot understand the result :

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

Anyone can tell me the answer, I think the three var point the same instance.

Answer 1

$instance = new SimpleClass(); // create instance
$assigned   =  $instance; // assign *identifier* to $assigned
$reference  =& $instance; // assign *reference* to $reference 

$instance->var = '$assigned will have this value';
$instance = null; // change $instance to null (as well as any variables that reference same)

Assigning via reference and identifier are different. From the manual:

One of the key-points of PHP5 OOP that is often mentioned is that "objects are passed by references by default". This is not completely true. This section rectifies that general thought using some examples.

A PHP reference is an alias, which allows two different variables to write to the same value. As of PHP5, an object variable doesn't contain the object itself as value anymore. It only contains an object identifier which allows object accessors to find the actual object. When an object is sent by argument, returned or assigned to another variable, the different variables are not aliases: they hold a copy of the identifier, which points to the same object.

Check out this answer for more info.