Passing by reference; why is the original object not changed?

Question

If objects are passed by reference in PHP5, then why $foo below doesn't change?

$foo = array(1, 2, 3);
$foo = (object)$foo;

$x = $foo;            // $x = &$foo makes $foo (5)!
$x = (object)array(5);

print_r($foo); // still 1,2,3

---

so:

Passing by reference not the same as assign.

then why $foo below is (100, 2, 3) ?

$foo = array('xxx' => 1, 'yyy' => 2, 'zzz' => 3);
$foo = (object)$foo;

$x = $foo;            
$x->xxx = 100;

print_r($foo);

Answer 1

The problem lies here:

$x = $foo;   
$x = (object)array(5);

On the first rule $x is referenced to $foo; editing $x wil also edit $foo;
(this is called "assign by reference", not "pass by reference" *1)

$x->myProperty= "Hi";

Will cause $foo to also have a property "myProperty".

But on the next line you reference $x to a new object.
Effectively unreferencing $x from $foo, all changes you make to $x won't propogate to $foo.

*1: When you call a function, the objects you pass to the functions are (in php5) "passed by reference"

Answer 2

Not only are objects passed by reference; they are also assigned by reference (which is what you're actually talking about):

An exception to the usual assignment by value behaviour within PHP occurs with objects, which are assigned by reference in PHP 5.

However, in your first example, you're performing a cast operation. This entails a copy:

If a value of any other type is converted to an object, a new instance of the stdClass built-in class is created.

Arrays have their own type in PHP, and are not objects; thus the above rule applies.

Answer 3

Passing by reference not the same as assign.