How do I create a copy of an object in PHP?


Question

It appears that in PHP objects are passed by reference. Even assignment operators do not appear to be creating a copy of the Object.

Here's a simple, contrived proof:

<?php

class A {
    public $b;
}


function set_b($obj) { $obj->b = "after"; }

$a = new A();
$a->b = "before";
$c = $a; //i would especially expect this to create a copy.

set_b($a);

print $a->b; //i would expect this to show 'before'
print $c->b; //i would ESPECIALLY expect this to show 'before'

?>

In both print cases I am getting 'after'

So, how do I pass $a to set_b() by value, not by reference?

1
143
7/12/2014 7:52:52 AM

Accepted Answer

In PHP 5+ objects are passed by reference. In PHP 4 they are passed by value (that's why it had runtime pass by reference, which became deprecated).

You can use the 'clone' operator in PHP5 to copy objects:

$objectB = clone $objectA;

Also, it's just objects that are passed by reference, not everything as you've said in your question...

244
10/9/2008 4:24:09 AM

The answers are commonly found in Java books.

  1. cloning: If you don't override clone method, the default behavior is shallow copy. If your objects have only primitive member variables, it's totally ok. But in a typeless language with another object as member variables, it's a headache.

  2. serialization/deserialization

$new_object = unserialize(serialize($your_object))

This achieves deep copy with a heavy cost depending on the complexity of the object.


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