PHP: 반환값

반환값

선택적인 return문을 사용하여 값을 돌려준다. 배열이나 객체를 포함하여 모든 타입을 돌려줄수있다. 이 구문에서 함수의 수행이 즉시 중단되고 현재 함수를 호출한 코드줄로 제어를 되돌린다. 자세한 정보는 return섹션을 참고할것.

Note:
If the return is omitted the value NULL will be returned.

Use of return

Example #1 return의 사용예


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

함수는 여러 값을 반환할 수 없습니다. 그러나 배열을 반환하게 해서 비슷한 결과를 얻을 수 있습니다.

Example #2 여러 값을 취하기 위해 배열을 돌려줌


<?php
function small_numbers()
{
    return array (0, 1, 2);
}
list ($zero, $one, $two) = small_numbers();
?>

함수에서 참조를 돌려주기위해서는, 함수 선언부와 돌려주는 변수값을 지정하는 곳에서 참조 연산자 & 를 사용해야 한다.

Example #3 함수에서 참조 돌려주기


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

$newref =& returns_reference();
?>

참조에 관한 자세한 정보는, 참조 표현섹션을 참고.

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.

예제

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));
?>

위 예제의 출력:

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));
?>

위 예제의 출력:

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());
?>

위 예제의 출력:

object(C)#1 (0) {
}

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User Contributed Notes 5 notes

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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.

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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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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.

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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

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user ¶

2 months ago


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