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?

Answer 1

According to the docs (http://ca3.php.net/language.oop5.cloning):

$a = clone $b;

Answer 2

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...

Answer 3

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.

Answer 4

According to previous comment, if you have another object as a member variable, do following:

class MyClass {
  private $someObject;

  public function __construct() {
    $this->someObject = new SomeClass();
  }

  public function __clone() {
    $this->someObject = clone $this->someObject;
  }

}

Now you can do cloning:

$bar = new MyClass();
$foo = clone $bar;