PHP: 返回值

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。

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用户贡献的备注 4 notes

down

rstaveley at seseit dot com ¶

14 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

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.

down

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.

down

nick at itomic.com ¶

21 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