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。
Example #1 return 的使用
<?php
function square($num)
{
return $num * $num;
}
echo square(4); // outputs '16'.
?>
函数不能返回多个值，但可以通过返回一个数组来得到类似的效果。
Example #2 返回一个数组以得到多个返回值
<?php
function small_numbers()
{
return array (0, 1, 2);
}
list ($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;
}
$newref =& returns_reference();
?>
有关引用的更多信息, 请查看引用的解释。
PHP 7 adds support for return type declarations. Similarly to argument type declarations, return type declarations specify the type of the value that will be returned from a function. The same types are available for return type declarations as are available for argument type declarations.
Strict typing also has an effect on return type declarations. In the default weak mode, returned values will be coerced to the correct type if they are not already of that type. In strong mode, the returned value must be of the correct type, otherwise a TypeError will be thrown.
Note:
When overriding a parent method, the child's method must match any return type declaration on the parent. If the parent doesn't define a return type, then the child method may do so.
Example #4 Basic return type declaration
<?php
function sum($a, $b): float {
return $a + $b;
}
// Note that a float will be returned.
var_dump(sum(1, 2));
?>
以上例程会输出：
float(3)
Example #5 Strict mode in action
<?php
declare(strict_types=1);
function sum($a, $b): int {
return $a + $b;
}
var_dump(sum(1, 2));
var_dump(sum(1, 2.5));
?>
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 Returning an object
<?php
class C {}
function getC(): C {
return new C;
}
var_dump(getC());
?>
以上例程会输出：
object(C)#1 (0) {
}
If a function/method parameter has a type declaration , then php compiler will check it the moment of invocation regardless if strict_types is set to 1 or 0 or not set at all.
However, php will check parameters types on build in php functions when strict_types is set to 1;
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.
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!
}
}
?>
I like this method of concatinating methods by returning $this. It makes my code more readable. If the returned value is not an object it will fail and you find mistakes easier.
<?php
class Dummy {
private $result;
function __construct()
{
$this->result =
$this
->setStuff('abc')
->generateResult()
;
}
function setStuff($value)
{
// do something
return $this;
}
function generateResult()
{
return ['the result'];
}
function getResult()
{
return $this->result;
}
}
?>
Note: the function does not have "alternative syntax" as if/endif, while/endwhile, and colon (:) here is used to define returning type and not to mark where the block statement begins.
Just a quick clarification on whether variables are passed by reference or not. Variables are always passed using a pointer, and if the variable is modified, it is copied and re-assigned. For example:
<?php
function byPointer($x) {
return $x / 3; // Does not modify or create a copy of $x
}
function copied($x) {
$x++; // At this point, creates a copy of $x to be used in local scope
return $x;
}
class Obj {
public function performAction() {}
public $y;
}
function objPointer(Obj $x) {
$x->performAction(); //works on $x
$x->y = '150'; //works on $x
$x = new Obj(); // Does not modify $x outside of function
$x->y = '250';
}
function objReference(Obj &$x) {
$x->performAction(); //works on $x
$x->y = '150'; //works on $x
$x = new Obj(); // Modifies original $x outside of function
$x->y = '250';
}
$x = new Obj();
$x->y = '10';
objPointer($x);
echo "Post Pointer: {$x->y}\n";
$x->y = '10';
objReference($x);
echo "Post Reference: {$x->y}\n";
?>
This will output:
Post Pointer: 150
Post Reference: 250
So make sure when writing functions that if you want to pass by reference you actually mean by reference, and not using standard PHP pointers
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
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";
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 $x;
}
echo myfunc1(). "\n". myfunc2(). "\n". myfunc3(). "\n". myfunc4(). "\n". myfunc5(). "\n";
?>
Only myfunc5() returns 'thingy' - the rest return 1.
Functions which return references, may return a NULL value. This is inconsistent with the fact that function parameters passed by reference can't be passed as NULL (or in fact anything which isnt a variable).
i.e.
<?php
function &testRet()
{
return NULL;
}
if (testRet() === NULL)
{
echo "NULL";
}
?>
parses fine and echoes NULL
In reference to the poster above, an additional (better?) way to return multiple values from a function is to use list(). For example:
function fn($a, $b)
{
# complex stuff
return array(
$a * $b,
$a + $b,
);
}
list($product, $sum) = fn(3, 4);
echo $product; # prints 12
echo $sum; # prints 7