Returning values

Values are returned by using the optional return statement. Any type may be returned, including arrays and objects. This causes the function to end its execution immediately and pass control back to the line from which it was called. See return for more information.

Note:

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

Example #1 Use of return

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

A function can not return multiple values, but similar results can be obtained by returning an array.

Example #2 Returning an array to get multiple values

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

To return a reference from a function, use the reference operator & in both the function declaration and when assigning the returned value to a variable:

Example #3 Returning a reference from a function

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

$newref =& returns_reference();
?>

For more information on references, please check out References Explained.

Коментарии

Автор:
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
2003-08-04 03:56:43
http://php5.kiev.ua/manual/ru/functions.returning-values.html
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.
2008-03-27 20:27:13
http://php5.kiev.ua/manual/ru/functions.returning-values.html
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->1;

$obj2 obj_inc_x($obj);
obj_inc_x($obj2);

print 
$obj->', ' $obj2->"\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";
2010-08-29 16:26:43
http://php5.kiev.ua/manual/ru/functions.returning-values.html
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 $fileNamestring $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.
2017-01-02 20:40:29
http://php5.kiev.ua/manual/ru/functions.returning-values.html

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