Php function return select

Php function return select

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.

Use of return

Example #1 Use of return

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

function small_numbers ()
return [ 0 , 1 , 2 ];
>
// Array destructuring will collect each member of the array individually
[ $zero , $one , $two ] = small_numbers ();

// Prior to 7.1.0, the only equivalent alternative is using list() construct
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

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

User Contributed Notes 10 notes

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:

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.

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.

# (1) Objects are always passed by reference and returned by reference

function obj_inc_x ( $obj ) $obj -> x ++;
return $obj ;
>

$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 ;
>

$x2 = scalar_inc_x ( $x );
scalar_inc_x ( $x2 );

# (3) You have to force pass by reference and return by reference on scalars

$x2 =& scalar_ref_inc_x ( $x ); # Need reference here as well as the function sig
scalar_ref_inc_x ( $x2 );

# (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

$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:

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.

With 7.1, these are possible yet;

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!
>
>
?>

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).

if ( testRet () === NULL )
echo «NULL» ;
>
?>

parses fine and echoes NULL

PHP 7 return types if specified can not return a null.
For example:
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

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.

You may specify child return type if there is no parent:

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 ;
>
>
?>

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.

Declaring a collection of objects as return type is not implemented and forbidden:
class Child <>

function getChilds (): Child [] return [(new Child ()), (new Child ())];
>

var_dump ( getChilds ());
// Returns: Parse error: syntax error, unexpected ‘[‘, expecting ‘ ?>

We have to use:
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:
function setChilds ( Child [] $childs )<>
// Not allowed

function setChilds (array $childs )<>
// Allowed
?>

Источник

Php function return select

Значения возвращаются при помощи необязательного оператора возврата. Возвращаемые значения могут быть любого типа, в том числе это могут быть массивы и объекты. Возврат приводит к завершению выполнения функции и передаче управления обратно к той строке кода, в которой данная функция была вызвана. Для получения более детальной информации ознакомьтесь с описанием return .

Замечание:

Если конструкция return не указана, то функция вернёт значение null .

Использование выражения return

Пример #1 Использование конструкции return

Функция не может возвращать несколько значений, но аналогичного результата можно добиться, возвращая массив.

Пример #2 Возврат нескольких значений в виде массива

function small_numbers ()
return [ 0 , 1 , 2 ];
>
// Деструктуризация массива будет собирать каждый элемент массива индивидуально
[ $zero , $one , $two ] = small_numbers ();

// До версии 7.1.0 единственной эквивалентной альтернативой было использование конструкции list().
list( $zero , $one , $two ) = small_numbers ();

Для того, чтобы функция возвращала результат по ссылке, вам необходимо использовать оператор & и при описании функции, и при присвоении переменной возвращаемого значения:

Пример #3 Возврат результата по ссылке

Для получения более детальной информации о ссылках обратитесь к разделу документации Подробно о ссылках.

User Contributed Notes 10 notes

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:

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.

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.

# (1) Objects are always passed by reference and returned by reference

function obj_inc_x ( $obj ) $obj -> x ++;
return $obj ;
>

$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 ;
>

$x2 = scalar_inc_x ( $x );
scalar_inc_x ( $x2 );

# (3) You have to force pass by reference and return by reference on scalars

$x2 =& scalar_ref_inc_x ( $x ); # Need reference here as well as the function sig
scalar_ref_inc_x ( $x2 );

# (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

$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:

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.

With 7.1, these are possible yet;

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!
>
>
?>

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).

if ( testRet () === NULL )
echo «NULL» ;
>
?>

parses fine and echoes NULL

PHP 7 return types if specified can not return a null.
For example:
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

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.

You may specify child return type if there is no parent:

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 ;
>
>
?>

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.

Declaring a collection of objects as return type is not implemented and forbidden:
class Child <>

function getChilds (): Child [] return [(new Child ()), (new Child ())];
>

var_dump ( getChilds ());
// Returns: Parse error: syntax error, unexpected ‘[‘, expecting ‘ ?>

We have to use:
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:
function setChilds ( Child [] $childs )<>
// Not allowed

function setChilds (array $childs )<>
// Allowed
?>

Источник

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