PHP 8.1.0 Released!

Dönen değerler

Değerler, isteğe bağlı return deyimi kullanılarak döndürülür. Diziler ve nesneler dahil herhangi bir tür döndürülebilir. Bu, işlevin çalıştırmayı hemen bitirmesine ve denetimi çağrıldığı satıra geri aktarmasına neden olur. Daha fazla bilgi için return işlevine bakın.

Bilginize:

return kullanılmazsa null değeri döndürülür.

Örnek 1 - return kullanımı

<?php
function square($num)
{
    return 
$num $num;
}
echo 
square(4);   // çıktısı: '16'.
?>

Bir işlev çok sayıda değer döndürebildiği gibi bir dizi döndürerek de aynı sonuç sağlanabilir.

Örnek 2 - Çok sayıda değeri bir diziyle döndürmek

<?php
function small_numbers()
{
    return (
012);
}

// Burada dizi üyeleri tek tek toplanmaktadır
[$zero$one$two] = small_numbers();

// 7.1.0 öncesinde, bu sadece list() oluşumu ile sağlanabilirdi
list($zero$one$two) = small_numbers();

?>

Bir işlevden gönderim döndürmek için hem işlev bildiriminde hem de dönen değer bir değişkene atandığında gönderim işleci & kullanılır:

Örnek 3 - Bir işlevden gönderim döndürmek

<?php
function &returns_reference()
{
    return 
$someref;
}
?>

$newref =& returns_reference();

Gönderimler hakkında daha fazla bilgi için Gönderimlerle İlgili Herşey bölümüne bakınız.

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User Contributed Notes 10 notes

up
31
ryan dot jentzsch at gmail dot com
4 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:

<?php
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.
up
31
rstaveley at seseit dot com
11 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";
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16
bgalloway at citycarshare dot org
13 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.
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9
nick at itomic.com
18 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
up
4
k-gun !! mail
4 years ago
With 7.1, these are possible yet;

<?php
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
ret_nullable() ?int {
    if (...) {
        return
123;
    } else {
        return
null; // MUST!
   
}
}
?>
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2
Berniev
3 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.
up
1
zored dot box at gmail dot com
3 years ago
You may specify child return type if there is no parent:

<?php

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

class
B extends A {
    public function
f ($a): int // + return type, OK
   
{
        return
1;
    }
}

class
C extends A {
    public function
f (int $a) // + argument type, WARNING
   
{
        return
1;
    }
}
?>
up
2
ryan dot jentzsch at gmail dot com
6 years ago
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
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-1
php(@)genjo(DOT)fr
2 years ago
Declaring a collection of objects as return type is not implemented and forbidden:
<?php
class Child{}

function
getChilds(): Child[]
{
    return [(new
Child()), (new Child())];
}

var_dump(getChilds());
// Returns:  Parse error: syntax error, unexpected '[', expecting '{'
?>

We have to use:
<?php
class Child{}

function
getChilds(): array
{
    return [(new
Child()), (new Child())];
}

var_dump(getChilds());
// Returns:
/*
array (size=2)
  0 =>
    object(Child)[168]
  1 =>
    object(Child)[398]
*/
?>

Idem for function parameter:
<?php
function setChilds(Child[] $childs){}
// Not allowed

function setChilds(array $childs){}
// Allowed
?>
up
-1
Vidmantas Maskoliunas
5 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.
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