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    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 array (    0, 1, 2);
    }

    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
    7 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
    30
    rstaveley at seseit dot com
    13 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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    18
    bgalloway at citycarshare dot org
    16 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.
    up
    9
    nick at itomic.com
    20 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
    3
    Berniev
    5 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
    3
    zored dot box at gmail dot com
    6 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
    5
    k-gun !! mail
    7 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!
            }
    }
    ?>
    up
    2
    ryan dot jentzsch at gmail dot com
    8 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
    up
    -3
    php(@)genjo(DOT)fr
    4 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
    -4
    Vidmantas Maskoliunas
    8 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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