What References Are
What References Are Not
References Explained
PHP Manual

## What References Do

PHP references allow you to make two variables to refer to the same content. Meaning, when you do:

`<?php
$a =& $b;
?>`

it means that $a and $b point to the same content.

> Note: $a and $b are completely equal here. $a is not pointing to $b or vice versa. $a and $b are pointing to the same place.

> Note: If arrays with references are copied, their values are not dereferenced. This is valid also for arrays passed by value to functions.

> Note: If you assign, pass, or return an undefined variable by reference, it will get created.

> Example #1 Using references with undefined variables

> `<?php
> function foo(&$var) { }
> foo($a); // $a is "created" and assigned to null
> $b = array();
> foo($b['b']);
> var_dump(array_key_exists('b', $b)); // bool(true)
> $c = new StdClass;
> foo($c->d);
> var_dump(property_exists($c, 'd')); // bool(true)
> ?>`

The same syntax can be used with functions that return references, and with the new operator (in PHP 4.0.4 and later):

`<?php
$bar =& new fooclass();
$foo =& find_var($bar);
?>`

Since PHP 5, new returns a reference automatically, so using =& in this context is deprecated and produces an E_STRICT message.

> Note: Not using the & operator causes a copy of the object to be made. If you use $this in the class it will operate on the current instance of the class. The assignment without & will copy the instance (i.e. the object) and $this will operate on the copy, which is not always what is desired. Usually you want to have a single instance to work with, due to performance and memory consumption issues.
> While you can use the @ operator to mute any errors in the constructor when using it as @new, this does not work when using the &new statement. This is a limitation of the Zend Engine and will therefore result in a parse error.

Warning

If you assign a reference to a variable declared global inside a function, the reference will be visible only inside the function. You can avoid this by using the $GLOBALS array.

Example #2 Referencing global variables inside functions

`<?php
$var1 = "Example variable";
$var2 = "";
function global_references($use_globals)
{
global $var1, $var2;
if (!$use_globals) {
$var2 =& $var1; // visible only inside the function
} else {
$GLOBALS["var2"] =& $var1; // visible also in global context
}
}
global_references(false);
echo "var2 is set to '$var2'\n"; // var2 is set to ''
global_references(true);
echo "var2 is set to '$var2'\n"; // var2 is set to 'Example variable'
?>`

Think about global $var; as a shortcut to $var =& $GLOBALS['var'];. Thus assigning another reference to $var only changes the local variable's reference.

> Note: If you assign a value to a variable with references in a foreach statement, the references are modified too.

> Example #3 References and foreach statement

> `<?php
> $ref = 0;
> $row =& $ref;
> foreach (array(1, 2, 3) as $row) {
> // do something
> }
> echo $ref; // 3 - last element of the iterated array
> ?>`

The second thing references do is to pass variables by reference. This is done by making a local variable in a function and a variable in the calling scope referencing the same content. Example:

`<?php
function foo(&$var)
{
$var++;
}
$a=5;
foo($a);
?>`

will make $a to be 6. This happens because in the function foo the variable $var refers to the same content as $a. For more information on this, read the passing by reference section.

The third thing references can do is return by reference.

What References Are
What References Are Not
References Explained
PHP Manual