PHP 7.0.6 Released

Returning References

Returning by reference is useful when you want to use a function to find to which variable a reference should be bound. Do not use return-by-reference to increase performance. The engine will automatically optimize this on its own. Only return references when you have a valid technical reason to do so. To return references, use this syntax:

<?php
class foo {
    public 
$value 42;

    public function &
getValue() {
        return 
$this->value;
    }
}

$obj = new foo;
$myValue = &$obj->getValue(); // $myValue is a reference to $obj->value, which is 42.
$obj->value 2;
echo 
$myValue;                // prints the new value of $obj->value, i.e. 2.
?>
In this example, the property of the object returned by the getValue function would be set, not the copy, as it would be without using reference syntax.

Note: Unlike parameter passing, here you have to use & in both places - to indicate that you want to return by reference, not a copy, and to indicate that reference binding, rather than usual assignment, should be done for $myValue.

Note: If you try to return a reference from a function with the syntax: return ($this->value); this will not work as you are attempting to return the result of an expression, and not a variable, by reference. You can only return variables by reference from a function - nothing else. Since PHP 4.4.0 in the PHP 4 branch, and PHP 5.1.0 in the PHP 5 branch, an E_NOTICE error is issued if the code tries to return a dynamic expression or a result of the new operator.

To use the returned reference, you must use reference assigment:

<?php
function &collector() {
  static 
$collection = array();
  return 
$collection;
}
$collection = &collector();
$collection[] = 'foo';
?>
To pass the returned reference to another function expecting a reference you can use this syntax:
<?php
function &collector() {
  static 
$collection = array();
  return 
$collection;
}
array_push(collector(), 'foo');
?>

Note: Note that array_push(&collector(), 'foo'); will not work, it results in a fatal error.

User Contributed Notes

Spad-XIII
8 years ago
a little addition to the example of pixel at minikomp dot com here below
<?php

   
function &func(){
        static
$static = 0;
       
$static++;
        return
$static;
    }

   
$var1 =& func();
    echo
"var1:", $var1; // 1
   
func();
   
func();
    echo
"var1:", $var1; // 3
   
$var2 = func(); // assignment without the &
   
echo "var2:", $var2; // 4
   
func();
   
func();
    echo
"var1:", $var1; // 6
   
echo "var2:", $var2; // still 4

?>
benjamin dot delespierre at gmail dot com
5 years ago
Keep in mind that returning by reference doesn't work with __callStatic:

<?php
class Test {
  private static
$_inst;
  public static function &
__callStatic ($name, $args) {
    if (!isset(static::
$_inst)){
      echo
"create";
      static::
$_inst = (object)"test";
   }
   return static::
$_inst;
}

var_dump($a = &Test::abc()); // prints 'create'
$a = null;
var_dump(Test::abc()); // doesn't prints and the instance still exists in Test::$_inst
?>
willem at designhulp dot nl
10 years ago
There is an important difference between php5 and php4 with references.

Lets say you have a class with a method called 'get_instance' to get a reference to an exsisting class and it's properties.

<?php
class mysql {
    function
get_instance(){
       
// check if object exsists
       
if(empty($_ENV['instances']['mysql'])){
           
// no object yet, create an object
           
$_ENV['instances']['mysql'] = new mysql;
        }
       
// return reference to object
       
$ref = &$_ENV['instances']['mysql'];
        return
$ref;
    }
}
?>

Now to get the exsisting object you can use
mysql::get_instance();

Though this works in php4 and in php5, but in php4 all data will be lost as if it is a new object while in php5 all properties in the object remain.
szymoncofalik at gmail dot com
5 years ago
Sometimes, you would like to return NULL with a function returning reference, to indicate the end of chain of elements. However this generates E_NOTICE. Here is little tip, how to prevent that:

<?php
class Foo {
   const
$nullGuard = NULL;
  
// ... some declarations and definitions
  
public function &next() {
     
// ...
     
if (!$end) return $bar;
      else return
$this->nullGuard;
   }
}
?>

by doing this you can do smth like this without notices:

<?php
$f
= new Foo();
// ...
while (($item = $f->next()) != NULL) {
// ...
}
?>

you may also use global variable:
global $nullGuard;
return $nullGuard;
obscvresovl at NOSPAM dot hotmail dot com
11 years ago
An example of returning references:

<?

$var = 1;
$num = NULL;

function &blah()
{
    $var =& $GLOBALS["var"]; # the same as global $var;
    $var++;
    return $var;
}

$num = &blah();

echo $num; # 2

blah();

echo $num; # 3

?>

Note: if you take the & off from the function, the second echo will be 2, because without & the var $num contains its returning value and not its returning reference.
spidgorny at gmail dot com
6 years ago
When returning reference to the object member which is instantiated inside the function, the object is destructed upon returning (which is a problem). It's easier to see the code:

<?php

class MemcacheArray {
    public
$data;

    ...

   
/**
     * Super-clever one line cache reading AND WRITING!
     * Usage $data = &MemcacheArray::getData(__METHOD__);
     * Hopefully PHP will know that $this->data is still used
     * and will call destructor after data changes.
     * Ooops, it's not the case.   
     *
     * @return unknown
     */
   
function &getData($file, $expire = 3600) {
       
$o = new MemcacheArray($file, $expire);
        return
$o->data;
    }
?>

Here, destructor is called upon return() and the reference becomes a normal variable.

My solution is to store objects in a pool until the final exit(), but I don't like it. Any other ideas?

<?php
   
protected static $instances = array();

    function &
getData($file, $expire = 3600) {
       
$o = new MemcacheArray($file, $expire);
       
self::$instances[$file] = $o; // keep object from destructing too early
       
return $o->data;
    }
?>
stanlemon at mac dot com
8 years ago
I haven't seen anyone note method chaining in PHP5.  When an object is returned by a method in PHP5 it is returned by default as a reference, and the new Zend Engine 2 allows you to chain method calls from those returned objects.  For example consider this code:

<?php

class Foo {

    protected
$bar;

    public function
__construct() {
       
$this->bar = new Bar();

        print
"Foo\n";
    }   
   
    public function
getBar() {
        return
$this->bar;
    }
}

class
Bar {

    public function
__construct() {
        print
"Bar\n";
    }
   
    public function
helloWorld() {
        print
"Hello World\n";
    }
}

function
test() {
    return new
Foo();
}

test()->getBar()->helloWorld();

?>

Notice how we called test() which was not on an object, but returned an instance of Foo, followed by a method on Foo, getBar() which returned an instance of Bar and finally called one of its methods helloWorld().  Those familiar with other interpretive languages (Java to name one) will recognize this functionality.  For whatever reason this change doesn't seem to be documented very well, so hopefully someone will find this helpful.
sandaimespaceman at gmail dot com
7 years ago
The &b() function returns a reference of $a in the global scope.

<?php
$a
= 0;
function &
b()
{
    global
$a;
    return
$a;
}
$c = &b();
$c++;
echo
"
\$a:
$a
\$b:
$c
"
?>

It outputs:

$a: 1 $b: 1
civilization28 at gmail dot com
1 year ago
Zayfod's example above is useful, but I feel that it needs more explanation. The point that should be made is that a parameter passed in by reference can be changed to reference something else, resulting in later changes to the local variable not affecting the passed in variable:

<?php

function    & func_b ()
{
   
$some_var = 2;
    return
$some_var;
}

function   
func_a (& $param)
{
   
# $param is 1 here
   
$param = & func_b();    # Here the reference is changed and
                                           # the "&" in "func_a (& $param)"
                                           # is no longer in effect at all.
    # $param is 2 here
   
$param++;    # Has no effect on $var.
}

$var = 1;
func_a($var);
# $var is still 1 here!!!    Because the reference was changed.

?>
Anonymous
2 years ago
I learned a painful lesson working with a class method that would pass by reference.  

In short, if you have a method in a class that is initialed with ampersand during declaration, do not use another ampersand when using the method as in &$this->method();

For example
<?php
class A {
    public function &
hello(){
        static
$a='';
        return
$a;
    }
    public function
bello(){
       
$b=&$this->hello();  // incorrect. Do not use ampersand.
       
$b=$this->hello();  // $b is a reference  to the static variable.
}
?>
php at thunder-2000 dot com
9 years ago
If you want to get a part of an array to manipulate, you can use this function

function &getArrayField(&$array,$path) {
  if (!empty($path)) {
    if (empty($array[$path[0]])) return NULL;
    else return getArrayField($array[$path[0]], array_slice($path, 1));
  } else {
    return $array;
  }
}

Use it like this:

$partArray =& getArrayField($GLOBALS,array("config","modul1"));

You can manipulate $partArray and the changes are also made with $GLOBALS.
rwruck
10 years ago
The note about using parentheses when returning references is only true if the variable you try to return does not already contain a reference.

<?php
// Will return a reference
function& getref1()
  {
 
$ref =& $GLOBALS['somevar'];
  return (
$ref);
  }

// Will return a value (and emit a notice)
function& getref2()
  {
 
$ref = 42;
  return (
$ref);
  }

// Will return a reference
function& getref3()
  {
  static
$ref = 42;
  return (
$ref);
  }
?>
Karim dot Khaled dot Wahby at outlook dot com
1 year ago
You can also return a function like this :
<?php
           
function ReturnFunc ($num, $num1)
            {
                if (
$num == 1)
                {
                   
// use is to inheret the variable $num1 from the calling function
                    // You can also Use global to refrance to global vars ;
                   
return function ($num2) use ($num1){return $num1+$num2;} ;
                }
                else if (
$num == 2)
                {
                    return function (
$num2) use ($num1){return $num1-$num2;} ;
                }
            }
           
            function
InvokeTheReturnedFunc($num,$num1, $num2)
            {
               
$func = ReturnFunc($num, $num1) ;
                echo
gettype($func) . '<br />'; // Object
               
return $func($num2) ;
            }
            echo
InvokeTheReturnedFunc(1, 5, 8) . '<br />'; // 13
           
echo InvokeTheReturnedFunc(2, 5, 8) . '<br />'; // -3
           
echo $asdasd ; // would generate a notice
  
?>
contact at infopol dot fr
12 years ago
A note about returning references embedded in non-reference arrays :

<?
$foo;

function bar () {
    global $foo;
    $return = array();
    $return[] =& $foo;
    return $return;
}

$foo = 1;
$foobar = bar();
$foobar[0] = 2;
echo $foo;
?>

results in "2" because the reference is copied (pretty neat).
hawcue at yahoo dot com
12 years ago
Be careful when using tinary operation condition?value1:value2

See the following code:

$a=1;
function &foo()
{
  global $a;
  return isset($a)?$a:null;
}
$b=&foo();
echo $b;   // shows 1
$b=2;
echo $a;   // shows 1 (not 2! because $b got a copy of $a)

To let $b be a reference to $a, use "if..then.." in the function.
zayfod at yahoo dot com
12 years ago
There is a small exception to the note on this page of the documentation. You do not have to use & to indicate that reference binding should be done when you assign to a value passed by reference the result of a function which returns by reference.

Consider the following two exaples:

<?php

function    & func_b ()
{
   
$some_var = 2;
    return
$some_var;
}

function   
func_a (& $param)
{
   
# $param is 1 here
   
$param = & func_b();
   
# $param is 2 here
}

$var = 1;
func_a($var);
# $var is still 1 here!!!

?>

The second example works as intended:

<?php

function    & func_b ()
{
   
$some_var = 2;
    return
$some_var;
}

function   
func_a (& $param)
{
   
# $param is 1 here
   
$param = func_b();
   
# $param is 2 here
}

$var = 1;
func_a($var);
# $var is 2 here as intended

?>

(Experienced with PHP 4.3.0)
pixel at minikomp dot com
8 years ago
<?php

   
function &func(){
        static
$static = 0;
       
$static++;
        return
$static;
    }

   
$var =& func();
    echo
$var; // 1
   
func();
   
func();
   
func();
   
func();
    echo
$var; // 5

?>
To Top