PHP - Manual: 返回值

2024-11-22

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。

注意:
如果省略了 return，则返回值为 null。

return 的使用

示例 #1 return 的使用


<?php
function square($num)
{
    return $num * $num;
}
echo square(4);   // 输出 '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。

示例 #2 返回一个数组以得到多个返回值


<?php
function small_numbers()
{
    return [0, 1, 2];
}
// 使用短数组语法将数组中的值赋给一组变量
[$zero, $one, $two] = small_numbers();

// 在 7.1.0 之前，唯一相等的选择是使用 list() 结构
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 &：

示例 #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。

add a note

User Contributed Notes 10 notes

down

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:

<?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.

down

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";

down

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:

<?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.

down

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

down

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.

down

k-gun !! mail ¶

5 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!
    }
}
?>

down

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

down

zored dot box at gmail dot com ¶

4 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;
    }
}
?>

down

-3

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.

down

-3

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
?>

add a note

官方地址：https://www.php.net/manual/en/functions.returning-values.php

有任何技术问题请点击这里网站运营推广招聘

IT PHP 编程语言 Linux 开发编程科技 Elasticsearch HTML/CSS/XML 网络 JAVA NoSQL 数据库面试 C/C++ Golang Git 算法操作系统正则表达式 Redis 互联网 MySql JavaScript 运维软件国际架构设计 Mac OS TCP/IP Excel Vim Windows Socket Oracle VR MongoDB Python 运营 MemCache 商业硬件电子娱乐设计 nginx 摄影游戏 WordPress HTTP 团建数码电器

php7 安装fileinfo扩展 preg_split — 通过一个正则表达式分隔字符串 php位值,解决 PHP 中 usort 在值相同时改变原始位置的问题 [PHP] inet_pton/inet_ntop IP地址转换函数 array_multisort() 多字段排序 PHPMailer设置utf8 PHPMailer字符集CharSet ADORecordSet对象 PHP 8.1 [鸟哥]PHP_INT_MIN 和 -9223372036854775808 adodb手册 opcache预加载 Composer的Packagist资源 mysql面试题 PHP获取指定函数定义在哪个文件中及行号 PHP7 mcrypt替代方案 Packagist / Composer 中国全量镜像 laravel框架查看sql语句方式 php中几种注释规范 php 修改图片大小 html转义/html编码/html解码

略微加速

PHP官方手册 - 互联网笔记

返回值

return 的使用

User Contributed Notes 10 notes