Php set variable to function return

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 null will be returned.

Use of return

Example #1 Use of return

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

function small_numbers()
{
    return [0, 1, 2];
}
// Array destructuring will collect each member of the array individually
[$zero, $one, $two] = small_numbers();// Prior to 7.1.0, the only equivalent alternative is using list() construct
list($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

function &returns_reference()
{
    return $someref;
}$newref =& returns_reference();
?>

For more information on references, please check out References Explained.

ryan dot jentzsch at gmail dot com ¶

5 years ago

PHP 7.1 allows for void and null return types by preceding the type declaration with a ? -- (e.g. function canReturnNullorString(): ?string)

However resource is not allowed as a return type:

function fileOpen(string $fileName, string $mode): resource
{
    $handle = fopen($fileName, $mode);
    if ($handle !== false)
    {
        return $handle;
    }
}$resourceHandle = fileOpen("myfile.txt", "r");
?>

Errors with:
Fatal error: Uncaught TypeError: Return value of fileOpen() must be an instance of resource, resource returned.

rstaveley at seseit dot com ¶

12 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.

# (1) Objects are always passed by reference and returned by referenceclass Obj {
    public $x;
}

function

bgalloway at citycarshare dot org ¶

14 years ago

Be careful about using "do this thing or die()" logic in your return lines.  It doesn't work as you'd expect:

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 ¶

19 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.

function &testRet()
{
    return NULL;
}

if (

k-gun !! mail ¶

5 years ago

With 7.1, these are possible yet;

function ret_void(): void {
    // do something but no return any value
    // if needs to break fn exec for any reason simply write return;
    if (...) {
        return; // break
        // return null; // even this NO!
    }$db->doSomething();
    // no need return call anymore
}

function

Berniev ¶

4 years ago

Be careful when introducing return types to your code.

Only one return type can be specified (but prefacing with ? allows null).

Return values of a type different to that specified are silently converted with sometimes perplexing results. These can be tedious to find and will need rewriting, along with calling code.

Declare strict types using "declare(strict_types=1);" and an error will be generated, saving much head-scratching.

ryan dot jentzsch at gmail dot com ¶

7 years ago

PHP 7 return types if specified can not return a null.
For example:
declare(strict_types=1);

function

zored dot box at gmail dot com ¶

4 years ago

You may specify child return type if there is no parent:

class A {
    public function f ($a)
    {
        return 1;
    }
}

class

Vidmantas Maskoliunas ¶

6 years ago

Note: the function does not have "alternative syntax" as if/endif, while/endwhile, and colon (:) here is used to define returning type and not to mark where the block statement begins.

php(@)genjo(DOT)fr ¶

2 years ago

Declaring a collection of objects as return type is not implemented and forbidden:
class Child{}

function