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Home/ Questions/Q 7178397
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Editorial Team
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Editorial Team
Asked: May 28, 20262026-05-28T16:54:48+00:00 2026-05-28T16:54:48+00:00

In the below example, any operation done by $instance2 and $instance3 modifies original object.

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In the below example, any operation done by $instance2 and $instance3 modifies original object.

My question is:

If a copy of an original object identifier and a reference to the original object identifier does same job, which one should be used in real applications?

What are the pros and cons of using a copy of object identifier and of using a reference to the object identifier?

I read the PHP manual but am unable to differentiate in terms of usage because both do the same job.

$instance1 = new test(1);
$instance2 = $instance1;
$instance3 =& $instance1;

//$instance1 -> original object identifier of the new object.
//$instance2 -> copy of object identifier $instance1
//$instance3 -> reference to the object identifier $instance1
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  1. Editorial Team
    Editorial Team
    2026-05-28T16:54:49+00:00Added an answer on May 28, 2026 at 4:54 pm

    $instance2 has a copy of the identifier to the object test. So, it contains the same as $instance1.
    $instance3 contains a reference to $instance1. The difference would be the following:

    $instance1 = new Test();
    $instance2 = $instance1;
    $instance3 = & $instance1;
    
    var_dump($instance1 instanceof Test); // True
    var_dump($instance2 instanceof Test); // True
    var_dump($instance3 instanceof Test); // True
    
    $instance3 = new AnotherTest();
    
    var_dump($instance1 instanceof AnotherTest); // True
    var_dump($instance2 instanceof AnotherTest); // False
    var_dump($instance3 instanceof AnotherTest); // True
    

    The same output would be returned if $instance1 was changed instead of $instance3.

    But if we did the following:

    $instance1 = new Test();
    $instance2 = $instance1;
    $instance3 = & $instance1;
    
    $instance2 = new AnotherTest();
    
    var_dump($instance1 instanceof AnotherTest); // False
    var_dump($instance2 instanceof AnotherTest); // True
    var_dump($instance3 instanceof AnotherTest); // False
    

    So:

    Modification of a variable which has been passed by reference or assigned by reference (using the & operand) or of the variable to which it references, modifies both, while modification of a copied variable modifies only the given variable.

    Still, you must remember that what $instance1 keeps is an identifier of the object, so:

    $instance1 = new StdClass();
    $instance2 = $instance1;
    
    $instance1->my_property = 1;
    var_dump($instance2->my_property); // Output: 1
    

    Hope it’s clearer now.

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