PHP/referencing

In PHP, a variable can be accessed either by value or by reference. All PHP variables are references; the question is what you're accessing – the location of the value or what's stored there. By default, you're accessing the value; referencing syntax allows you to refer to the location instead. Unlike with a pointer, however, the location's address is never accessible to PHP code; it can only be copied.

Concepts

Each variable in PHP is a reference to a storage location; the details of this are generally hidden. When one variable is assigned a reference to another (rather than just the value) they then both refer to the same storage location. This storage location will continue to exist as long as at least one variable is using it.

References do allow some of the functionality provided by pointers in other languages (like C/C++), but unlike pointers they are always de-referenced from the standpoint of the PHP code. A better analogy (the PHP manual agrees) is to think of them like hard file links in Unix. Conceptually, $b = &$a means "$b and $a now point to the same content", not "$b is a pointer to $a's content".

Clarification

There are no reference types (you can't declare, for example, &array or &int variables or return values); there are only reference operations -- and the reference operation is basically a qualifier (indicated by &) on a variable-access.

This is seemingly at odds with the fact that functions which return references are declared using the "&" prefix on the function name rather than on the return variable:

// THIS:
function &Ref(&$a) { $b=$a; return $b; }
// NOT THIS
function Ref(&$a) { $b=$a; return &$b; }
// OR THIS
function &Ref($a) { $b=$a; return $b; }
// AND DON'T FORGET
$y =& Ref(); // not $y = Ref();

Basics

To return a reference instead of a value, use the & operator when referring to the source variable:

$vtest = 'test value';
$vref = &$vtest;
echo "VTEST: ";
var_dump($vtest);
echo '<br>';
echo "VREF: ";
var_dump($vref);
echo '<br>';
$vref = 'another test';
echo "VTEST: ";
var_dump($vtest);
echo '<br>';

Output:

VTEST: string(10) "test value"
VREF: string(10) "test value"
VTEST: string(12) "another test"

This illustrates some key points:

• The var_dump() function can't tell the difference between a variable that was given a reference and one that was given a value.
  • ...because there isn't one, actually.
• Writing a value to a variable that contains a reference overwrites the value of the reference target, rather than replacing the reference.
  • This is kind of essential to how references work: spooky action at a distance, basically.
  • This also implies that the only way to change a reference-holding variable directly is to assign it a new reference or unset() it.

Links

• References Explained
• Assignment Operators: Assignment by Reference
• Function arguments: Passing arguments by reference
• Returning values: see "Example #3 Returning a reference from a function"
• Objects and references: they're not handled quite the same
  • short explanation: objects are looked up in a different table from references