Are PHP5 objects passed by reference? [duplicate]

Objects are passed (and assigned) by reference. No need to use address of operator.

Granted what I typed is an oversimplification but will suit your purposes. The documentation states:

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.

For a more detailed explanation (explains the oversimplification as well as identifiers) check out this answer.

From the PHP manual:

You can pass a variable by reference to a function so the function can modify the variable. The syntax is as follows:

<?php
function foo(&$var)
{
    $var++;
}

$a=5;
foo($a);
// $a is 6 here
?>

Note: There is no reference sign on a function call - only on function definitions. Function definitions alone are enough to correctly pass the argument by reference. As of PHP 5.3.0, you will get a warning saying that "call-time pass-by-reference" is deprecated when you use & in foo(&$a);.

And from What's New in PHP5:

In PHP 5, the infrastructure of the object model was rewritten to work with object handles. Unless you explicitly clone an object by using the clone keyword you will never create behind the scene duplicates of your objects. In PHP 5, there is neither a need to pass objects by reference nor assigning them by reference

So therefore the only time you need to use the function foo(&$var) syntax is if $var might not be an instance of a class.