User-defined functions
Returning values
Functions
PHP Manual

## Function arguments

Information may be passed to functions via the argument list, which is a comma-delimited list of expressions. The arguments are evaluated from left to right.

PHP supports passing arguments by value (the default), passing by reference, and default argument values. Variable-length argument lists are also supported.

Example #1 Passing arrays to functions

`<?php
function takes_array($input)
{
echo "$input[0] + $input[1] = ", $input[0]+$input[1];
}
?>`

### Passing arguments by reference

By default, function arguments are passed by value (so that if the value of the argument within the function is changed, it does not get changed outside of the function). To allow a function to modify its arguments, they must be passed by reference.

To have an argument to a function always passed by reference, prepend an ampersand (&) to the argument name in the function definition:

Example #2 Passing function parameters by reference

`<?php
function add_some_extra(&$string)
{
$string .= 'and something extra.';
}
$str = 'This is a string, ';
add_some_extra($str);
echo $str; // outputs 'This is a string, and something extra.'
?>`

### Default argument values

A function may define C++-style default values for scalar arguments as follows:

Example #3 Use of default parameters in functions

`<?php
function makecoffee($type = "cappuccino")
{
return "Making a cup of $type.\n";
}
echo makecoffee();
echo makecoffee(null);
echo makecoffee("espresso");
?>`

The above example will output:

    Making a cup of cappuccino.
    Making a cup of .
    Making a cup of espresso.

PHP also allows the use of arrays and the special type `NULL` as default values, for example:

Example #4 Using non-scalar types as default values

`<?php
function makecoffee($types = array("cappuccino"), $coffeeMaker = NULL)
{
$device = is_null($coffeeMaker) ? "hands" : $coffeeMaker;
return "Making a cup of ".join(", ", $types)." with $device.\n";
}
echo makecoffee();
echo makecoffee(array("cappuccino", "lavazza"), "teapot");
?>`

The default value must be a constant expression, not (for example) a variable, a class member or a function call.

Note that when using default arguments, any defaults should be on the right side of any non-default arguments; otherwise, things will not work as expected. Consider the following code snippet:

Example #5 Incorrect usage of default function arguments

`<?php
function makeyogurt($type = "acidophilus", $flavour)
{
return "Making a bowl of $type $flavour.\n";
}
echo makeyogurt("raspberry"); // won't work as expected
?>`

The above example will output:

    Warning: Missing argument 2 in call to makeyogurt() in
    /usr/local/etc/httpd/htdocs/phptest/functest.html on line 41
    Making a bowl of raspberry .

Now, compare the above with this:

Example #6 Correct usage of default function arguments

`<?php
function makeyogurt($flavour, $type = "acidophilus")
{
return "Making a bowl of $type $flavour.\n";
}
echo makeyogurt("raspberry"); // works as expected
?>`

The above example will output:

    Making a bowl of acidophilus raspberry.

> Note: As of PHP 5, arguments that are passed by reference may have a default value.

### Type declarations

> Note:

> Type declarations were also known as type hints in PHP 5.

Type declarations allow functions to specify parameters as certain types. If the given value is of the incorrect type, then an error is generated: in PHP 5, this will be a recoverable fatal error, while PHP 7 will throw a TypeError exception.

To specify a type declaration, the type name should be added before the parameter name. The declaration can be made to accept `NULL` values if the default value of the parameter is set to `NULL`.

#### Valid types

Type | Description | Minimum PHP version
--- | --- | ---
Class/interface name | The parameter must be an instanceof the given class or interface name. | PHP 5.0.0
self | The parameter must be an instanceof the same class as the one the method is defined on. This can only be used on class and instance methods. | PHP 5.0.0
array | The parameter must be an array. | PHP 5.1.0
callable | The parameter must be a valid callable. | PHP 5.4.0
bool | The parameter must be a boolean value. | PHP 7.0.0
float | The parameter must be a floating point number. | PHP 7.0.0
int | The parameter must be an integer. | PHP 7.0.0
string | The parameter must be a string. | PHP 7.0.0

Warning

Aliases for the above scalar types are not supported. Instead, they are treated as class or interface names. For example, using boolean as a parameter or return type will require an argument or return value that is an instanceof the class or interface boolean, rather than of type bool:

`<?php
function test(boolean $param) {}
test(true);
?>`

The above example will output:

     Fatal error: Uncaught TypeError: Argument 1 passed to test() must be an instance of boolean, boolean given, called in - on line 1 and defined in -:1

#### Examples

Example #7 Basic class type declaration

`<?php
class C {}
class D extends C {}
// This doesn't extend C.
class E {}
function f(C $c) {
echo get_class($c)."\n";
}
f(new C);
f(new D);
f(new E);
?>`

The above example will output:

    C
    D

    Fatal error: Uncaught TypeError: Argument 1 passed to f() must be an instance of C, instance of E given, called in - on line 14 and defined in -:8
    Stack trace:
    #0 -(14): f(Object(E))
    #1 {main}
      thrown in - on line 8

Example #8 Basic interface type declaration

`<?php
interface I { public function f(); }
class C implements I { public function f() {} }
// This doesn't implement I.
class E {}
function f(I $i) {
echo get_class($i)."\n";
}
f(new C);
f(new E);
?>`

The above example will output:

    C

    Fatal error: Uncaught TypeError: Argument 1 passed to f() must implement interface I, instance of E given, called in - on line 13 and defined in -:8
    Stack trace:
    #0 -(13): f(Object(E))
    #1 {main}
      thrown in - on line 8

Example #9 Nullable type declaration

`<?php
class C {}
function f(C $c = null) {
var_dump($c);
}
f(new C);
f(null);
?>`

The above example will output:

    object(C)#1 (0) {
    }
    NULL

#### Strict typing

By default, PHP will coerce values of the wrong type into the expected scalar type if possible. For example, a function that is given an integer for a parameter that expects a string will get a variable of type string.

It is possible to enable strict mode on a per-file basis. In strict mode, only a variable of exact type of the type declaration will be accepted, or a TypeError will be thrown. The only exception to this rule is that an integer may be given to a function expecting a float.

To enable strict mode, the declare statement is used with the strict_types declaration:

Caution

Enabling strict mode will also affect return type declarations.

> Note:

> Strict typing applies to function calls made from within the file with strict typing enabled, not to the functions declared within that file. If a file without strict typing enabled makes a call to a function that was defined in a file with strict typing, the caller's preference (weak typing) will be respected, and the value will be coerced.

> Note:

> Strict typing is only defined for scalar type declarations, and as such, requires PHP 7.0.0 or later, as scalar type declarations were added in that version.

Example #10 Strict typing

`<?php
declare(strict_types=1);
function sum(int $a, int $b) {
return $a + $b;
}
var_dump(sum(1, 2));
var_dump(sum(1.5, 2.5));
?>`

The above example will output:

    int(3)

    Fatal error: Uncaught TypeError: Argument 1 passed to sum() must be of the type integer, float given, called in - on line 9 and defined in -:4
    Stack trace:
    #0 -(9): sum(1.5, 2.5)
    #1 {main}
      thrown in - on line 4

Example #11 Weak typing

`<?php
function sum(int $a, int $b) {
return $a + $b;
}
var_dump(sum(1, 2));
// These will be coerced to integers: note the output below!
var_dump(sum(1.5, 2.5));
?>`

The above example will output:

    int(3)
    int(3)

Example #12 Catching TypeError

`<?php
declare(strict_types=1);
function sum(int $a, int $b) {
return $a + $b;
}
try {
var_dump(sum(1, 2));
var_dump(sum(1.5, 2.5));
} catch (TypeError $e) {
echo 'Error: '.$e->getMessage();
}
?>`

The above example will output:

    int(3)
    Error: Argument 1 passed to sum() must be of the type integer, float given, called in - on line 10

### Variable-length argument lists

PHP has support for variable-length argument lists in user-defined functions. This is implemented using the ... token in PHP 5.6 and later, and using the func_num_args(), func_get_arg(), and func_get_args() functions in PHP 5.5 and earlier.

#### ... in PHP 5.6+

In PHP 5.6 and later, argument lists may include the ... token to denote that the function accepts a variable number of arguments. The arguments will be passed into the given variable as an array; for example:

Example #13 Using ... to access variable arguments

`<?php
function sum(...$numbers) {
$acc = 0;
foreach ($numbers as $n) {
$acc += $n;
}
return $acc;
}
echo sum(1, 2, 3, 4);
?>`

The above example will output:

    10

You can also use ... when calling functions to unpack an array or Traversable variable or literal into the argument list:

Example #14 Using ... to provide arguments

`<?php
function add($a, $b) {
return $a + $b;
}
echo add(...[1, 2])."\n";
$a = [1, 2];
echo add(...$a);
?>`

The above example will output:

    3
    3

You may specify normal positional arguments before the ... token. In this case, only the trailing arguments that don't match a positional argument will be added to the array generated by ....

It is also possible to add a type hint before the ... token. If this is present, then all arguments captured by ... must be objects of the hinted class.

Example #15 Type hinted variable arguments

`<?php
function total_intervals($unit, DateInterval ...$intervals) {
$time = 0;
foreach ($intervals as $interval) {
$time += $interval->$unit;
}
return $time;
}
$a = new DateInterval('P1D');
$b = new DateInterval('P2D');
echo total_intervals('d', $a, $b).' days';
// This will fail, since null isn't a DateInterval object.
echo total_intervals('d', null);
?>`

The above example will output:

    3 days
    Catchable fatal error: Argument 2 passed to total_intervals() must be an instance of DateInterval, null given, called in - on line 14 and defined in - on line 2

Finally, you may also pass variable arguments by reference by prefixing the ... with an ampersand (&).

#### Older versions of PHP

No special syntax is required to note that a function is variadic; however access to the function's arguments must use func_num_args(), func_get_arg() and func_get_args().

The first example above would be implemented as follows in PHP 5.5 and earlier:

Example #16 Accessing variable arguments in PHP 5.5 and earlier

`<?php
function sum() {
$acc = 0;
foreach (func_get_args() as $n) {
$acc += $n;
}
return $acc;
}
echo sum(1, 2, 3, 4);
?>`

The above example will output:

    10

User-defined functions
Returning values
Functions
PHP Manual