The PHP Online Conference 2021

返回值

值通过使用可选的返回语句返回。可以返回包括数组和对象的任意类型。返回语句会立即中止函数的运行,并且将控制权交回调用该函数的代码行。更多信息见 return

Note:

如果省略了 return,则返回值为 NULL

return 的使用

Example #1 return 的使用

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

函数不能返回多个值,但可以通过返回一个数组来得到类似的效果。

Example #2 返回一个数组以得到多个返回值

<?php
function small_numbers()
{
    return array (
012);
}
list (
$zero$one$two) = small_numbers();
?>

从函数返回一个引用,必须在函数声明和指派返回值给一个变量时都使用引用运算符 &:

Example #3 从函数返回一个引用

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

$newref =& returns_reference();
?>

有关引用的更多信息, 请查看引用的解释

返回值类型声明

PHP 7 增加了对返回值类型声明的支持。 就如 类型声明一样, 返回值类型声明将指定该函数返回值的类型。同样,返回值类型声明也与 有效类型 中可用的参数类型声明一致。

严格类型 也会影响返回值类型声明。在默认的弱模式中,如果返回值与返回值的类型不一致,则会被强制转换为返回值声明的类型。在强模式中,返回值的类型必须正确,否则将会抛出一个TypeError异常.

As of PHP 7.1.0, return values can be marked as nullable by prefixing the type name with a question mark (?). This signifies that the function returns either the specified type or NULL.

Note:

当覆盖一个父类方法时,子类方法的返回值类型声明必须与父类一致。如果父类方法没有定义返回类型,那么子类方法可以定义任意的返回值类型声明。

范例

Example #4 基础返回值类型声明

<?php
function sum($a$b): float {
    return 
$a $b;
}

// Note that a float will be returned.
var_dump(sum(12));
?>

以上例程会输出:

float(3)

Example #5 严格模式下执行

<?php
declare(strict_types=1);

function 
sum($a$b): int {
    return 
$a $b;
}

var_dump(sum(12));
var_dump(sum(12.5));
?>

以上例程会输出:

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 返回一个对象

<?php
class {}

function 
getC(): {
    return new 
C;
}

var_dump(getC());
?>

以上例程会输出:

object(C)#1 (0) {
}

Example #7 Nullable return type declaration (as of PHP 7.1.0)

<?php
function get_item(): ?string {
    if (isset(
$_GET['item'])) {
        return 
$_GET['item'];
    } else {
        return 
null;
    }
}
?>
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User Contributed Notes 10 notes

up
32
ryan dot jentzsch at gmail dot com
3 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
28
rstaveley at seseit dot com
10 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";
up
19
bgalloway at citycarshare dot org
12 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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8
k-gun !! mail
3 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
3
php(@)genjo(DOT)fr
1 year 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
?>
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3
Berniev
2 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
2 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;
    }
}
?>
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9
nick at itomic.com
17 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
ryan dot jentzsch at gmail dot com
5 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
0
Vidmantas Maskoliunas
4 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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