PHP 7 return types if specified can not return a null.
For example:
<?php
declare(strict_types=1);
function add2ints(int $x, int $y):int
{
$z = $x + $y;
if ($z===0)
{
return null;
}
return $z;
}
$a = add2ints(3, 4);
echo is_null($a) ? 'Null' : $a;
$b = add2ints(-2, 2);
echo is_null($b) ? 'Null' : $b;
exit();
Output:
7
Process finished with exit code 139
Returning values
Values are returned by using the optional return statement. Any type may be returned, including arrays and objects. This causes the function to end its execution immediately and pass control back to the line from which it was called. See return for more information.
Note:
If the return is omitted the value
NULLwill be returned.
Use of return
Example #1 Use of return
<?php
function square($num)
{
return $num * $num;
}
echo square(4); // outputs '16'.
?>
A function can not return multiple values, but similar results can be obtained by returning an array.
Example #2 Returning an array to get multiple values
<?php
function small_numbers()
{
return array (0, 1, 2);
}
list ($zero, $one, $two) = small_numbers();
?>
To return a reference from a function, use the reference operator & in both the function declaration and when assigning the returned value to a variable:
Example #3 Returning a reference from a function
<?php
function &returns_reference()
{
return $someref;
}
$newref =& returns_reference();
?>
For more information on references, please check out References Explained.
Return type declarations
PHP 7 adds support for return type declarations. Similarly to argument type declarations, return type declarations specify the type of the value that will be returned from a function. The same types are available for return type declarations as are available for argument type declarations.
Strict typing also has an effect on return type declarations. In the default weak mode, returned values will be coerced to the correct type if they are not already of that type. In strong mode, the returned value must be of the correct type, otherwise a TypeError will be thrown.
Note:
When overriding a parent method, the child's method must match any return type declaration on the parent. If the parent doesn't define a return type, then the child method may do so.
Examples
Example #4 Basic return type declaration
<?php
function sum($a, $b): float {
return $a + $b;
}
// Note that a float will be returned.
var_dump(sum(1, 2));
?>
The above example will output:
float(3)
Example #5 Strict mode in action
<?php
declare(strict_types=1);
function sum($a, $b): int {
return $a + $b;
}
var_dump(sum(1, 2));
var_dump(sum(1, 2.5));
?>
The above example will output:
int(3)
Fatal error: Uncaught TypeError: Return value of sum() must be of the type integer, float returned in - on line 5 in -:5
Stack trace:
#0 -(9): sum(1, 2.5)
#1 {main}
thrown in - on line 5
Example #6 Returning an object
<?php
class C {}
function getC(): C {
return new C;
}
var_dump(getC());
?>
The above example will output:
object(C)#1 (0) {
}
add a note
User Contributed Notes 8 notes
ryan dot jentzsch at gmail dot com ¶
3 months ago
rstaveley at seseit dot com ¶
5 years ago
Developers with a C background may expect pass by reference semantics for arrays. It may be surprising that pass by value is used for arrays just like scalars. Objects are implicitly passed by reference.
<?php
# (1) Objects are always passed by reference and returned by reference
class Obj {
public $x;
}
function obj_inc_x($obj) {
$obj->x++;
return $obj;
}
$obj = new Obj();
$obj->x = 1;
$obj2 = obj_inc_x($obj);
obj_inc_x($obj2);
print $obj->x . ', ' . $obj2->x . "\n";
# (2) Scalars are not passed by reference or returned as such
function scalar_inc_x($x) {
$x++;
return $x;
}
$x = 1;
$x2 = scalar_inc_x($x);
scalar_inc_x($x2);
print $x . ', ' . $x2 . "\n";
# (3) You have to force pass by reference and return by reference on scalars
function &scalar_ref_inc_x(&$x) {
$x++;
return $x;
}
$x = 1;
$x2 =& scalar_ref_inc_x($x); # Need reference here as well as the function sig
scalar_ref_inc_x($x2);
print $x . ', ' . $x2 . "\n";
# (4) Arrays use pass by value sematics just like scalars
function array_inc_x($array) {
$array{'x'}++;
return $array;
}
$array = array();
$array['x'] = 1;
$array2 = array_inc_x($array);
array_inc_x($array2);
print $array['x'] . ', ' . $array2['x'] . "\n";
# (5) You have to force pass by reference and return by reference on arrays
function &array_ref_inc_x(&$array) {
$array{'x'}++;
return $array;
}
$array = array();
$array['x'] = 1;
$array2 =& array_ref_inc_x($array); # Need reference here as well as the function sig
array_ref_inc_x($array2);
print $array['x'] . ', ' . $array2['x'] . "\n";
bgalloway at citycarshare dot org ¶
7 years ago
Be careful about using "do this thing or die()" logic in your return lines. It doesn't work as you'd expect:
<?php
function myfunc1() {
return('thingy' or die('otherthingy'));
}
function myfunc2() {
return 'thingy' or die('otherthingy');
}
function myfunc3() {
return('thingy') or die('otherthingy');
}
function myfunc4() {
return 'thingy' or 'otherthingy';
}
function myfunc5() {
$x = 'thingy' or 'otherthingy'; return $x;
}
echo myfunc1(). "\n". myfunc2(). "\n". myfunc3(). "\n". myfunc4(). "\n". myfunc5(). "\n";
?>
Only myfunc5() returns 'thingy' - the rest return 1.
nick at itomic.com ¶
12 years ago
Functions which return references, may return a NULL value. This is inconsistent with the fact that function parameters passed by reference can't be passed as NULL (or in fact anything which isnt a variable).
i.e.
<?php
function &testRet()
{
return NULL;
}
if (testRet() === NULL)
{
echo "NULL";
}
?>
parses fine and echoes NULL
Nathan Salter ¶
5 days ago
Just a quick clarification on whether variables are passed by reference or not. Variables are always passed using a pointer, and if the variable is modified, it is copied and re-assigned. For example:
<?php
function byPointer($x) {
return $x / 3; // Does not modify or create a copy of $x
}
function copied($x) {
$x++; // At this point, creates a copy of $x to be used in local scope
return $x;
}
class Obj {
public function performAction() {}
public $y;
}
function objPointer(Obj $x) {
$x->performAction(); //works on $x
$x->y = '150'; //works on $x
$x = new Obj(); // Does not modify $x outside of function
$x->y = '250';
}
function objReference(Obj &$x) {
$x->performAction(); //works on $x
$x->y = '150'; //works on $x
$x = new Obj(); // Modifies original $x outside of function
$x->y = '250';
}
$x = new Obj();
$x->y = '10';
objPointer($x);
echo "Post Pointer: {$x->y}\n";
$x->y = '10';
objReference($x);
echo "Post Reference: {$x->y}\n";
?>
This will output:
Post Pointer: 150
Post Reference: 250
So make sure when writing functions that if you want to pass by reference you actually mean by reference, and not using standard PHP pointers
ian at NO_SPAM dot verteron dot net ¶
12 years ago
In reference to the poster above, an additional (better?) way to return multiple values from a function is to use list(). For example:
function fn($a, $b)
{
# complex stuff
return array(
$a * $b,
$a + $b,
);
}
list($product, $sum) = fn(3, 4);
echo $product; # prints 12
echo $sum; # prints 7
Anonymous ¶
5 years ago
As of at least PHP 5.3, a function or class method returning an object acts like an object.
<?php
class A {
function test() {
echo "Yay!";
}
}
function get_obj() {
return new A();
}
get_obj()->test(); // "Yay!"
?>
Sorry, still doesn't work with arrays. Ie <?php echo get_array()[1]; ?> fails.
ahmad at myandmyhost dot comze dot com ¶
1 year ago
we can return array using foreach
<?php
function my_func2()
{
return array("Name"=>"Ahmad Sayeed","Email"=>"[email protected]");
}
$ar = my_func2();
foreach($ar as $key=>$value)
{echo $key." , ".$value,"<br />";}
?>
