Support for type hinting array parameters (and return values after PHP 7.1) was added in PHP 5.1 with the keyword array. Any arrays of any dimensions and types, as well as empty arrays, are valid values.

Support for type hinting callables was added in PHP 5.4. Any value that is_callable() is valid for parameters and return values hinted callable, i.e. Closure objects, function name strings and array(class_name|object, method_name).

If a typo occurs in the function name such that it is not is_callable(), a less obvious error message would be displayed:

Fatal error: Uncaught TypeError: Argument 1 passed to foo() must be of the type callable, string/array given

function foo(callable $c) {}
foo("count"); // valid
foo("Phar::running"); // valid
foo(["Phar", "running"); // valid
foo([new ReflectionClass("stdClass"), "getName"]); // valid
foo(function() {}); // valid

foo("no_such_function"); // callable expected, string given

Nonstatic methods can also be passed as callables in static format, resulting in a deprecation warning and level E_STRICT error in PHP 7 and 5 respectively.

Method visibility is taken into account. If the context of the method with the callable parameter does not have access to the callable provided, it will end up as if the method does not exist.

class Foo{
  private static function f(){
    echo "Good" . PHP_EOL;
  }

  public static function r(callable $c){
    $c();
  }
}

function r(callable $c){}

Foo::r(["Foo", "f"]);
r(["Foo", "f"]);

Output:

Fatal error: Uncaught TypeError: Argument 1 passed to r() must be callable, array given

Support for type hinting scalar types was added in PHP 7. This means that we gain type hinting support for booleans, integers, floats and strings.

<?php

function add(int $a, int $b) {
    return $a + $b;
}

var_dump(add(1, 2)); // Outputs "int(3)"

By default, PHP will attempt to cast any provided argument to match its type hint. Changing the call to add(1.5, 2) gives exactly the same output, since the float 1.5 was cast to int by PHP.

To stop this behavior, one must add declare(strict_types=1); to the top of every PHP source file that requires it.

<?php

declare(strict_types=1);

function add(int $a, int $b) {
    return $a + $b;
}

var_dump(add(1.5, 2));

The above script now produces a fatal error:

Fatal error: Uncaught TypeError: Argument 1 passed to add() must be of the type integer, float given

An Exception: Special Types

Some PHP functions may return a value of type resource. Since this is not a scalar type, but a special type, it is not possible to type hint it.

As an example, curl_init() will return a resource, as well as fopen(). Of course, those two resources aren't compatible to each other. Because of that, PHP 7 will always throw the following TypeError when type hinting resource explicitly:

TypeError: Argument 1 passed to sample() must be an instance of resource, resource given

Type hinting for classes and interfaces was added in PHP 5.

Class type hint

<?php

class Student
{
    public $name = 'Chris';
}

class School
{
    public $name = 'University of Edinburgh';
}

function enroll(Student $student, School $school)
{
    echo $student->name . ' is being enrolled at ' . $school->name;
}

$student = new Student();
$school = new School();

enroll($student, $school);

The above script outputs:

Chris is being enrolled at University of Edinburgh

Interface type hint

<?php

interface Enrollable {};
interface Attendable {};

class Chris implements Enrollable
{
    public $name = 'Chris';
}

class UniversityOfEdinburgh implements Attendable
{
    public $name = 'University of Edinburgh';
}

function enroll(Enrollable $enrollee, Attendable $premises)
{
    echo $enrollee->name . ' is being enrolled at ' . $premises->name;
}

$chris = new Chris();
$edinburgh = new UniversityOfEdinburgh();

enroll($chris, $edinburgh);

The above example outputs the same as before:

Chris is being enrolled at University of Edinburgh

Self type hints

The self keyword can be used as a type hint to indicate that the value must be an instance of the class that declares the method.

Parameters

Nullable type hint was added in PHP 7.1 using the ? operator before the type hint.

function f(?string $a) {}
function g(string $a) {}

f(null); // valid
g(null); // TypeError: Argument 1 passed to g() must be of the type string, null given

Before PHP 7.1, if a parameter has a type hint, it must declare a default value null to accept null values.

function f(string $a = null) {}
function g(string $a) {}

f(null); // valid
g(null); // TypeError: Argument 1 passed to g() must be of the type string, null given

Return values

In PHP 7.0, functions with a return type must not return null.

In PHP 7.1, functions can declare a nullable return type hint. However, the function must still return null, not void (no/empty return statements).

function f() : ?string {
    return null;
}

function g() : ?string {}
function h() : ?string {}

f(); // OK
g(); // TypeError: Return value of g() must be of the type string or null, none returned
h(); // TypeError: Return value of h() must be of the type string or null, none returned

Since PHP objects don't inherit from any base class (including stdClass), there is no support for type hinting a generic object type.

For example, the below will not work.

<?php

function doSomething(object $obj) {
    return $obj;
}

class ClassOne {}
class ClassTwo {}

$classOne= new ClassOne();
$classTwo= new ClassTwo();

doSomething($classOne);
doSomething($classTwo);

And will throw a fatal error:

Fatal error: Uncaught TypeError: Argument 1 passed to doSomething() must be an instance of object, instance of OperationOne given

A workaround to this is to declare a degenerate interface that defines no methods, and have all of your objects implement this interface.

<?php

interface Object {}

function doSomething(Object $obj) {
    return $obj;
}

class ClassOne implements Object {}
class ClassTwo implements Object {}

$classOne = new ClassOne();
$classTwo = new ClassTwo();

doSomething($classOne);
doSomething($classTwo);

In PHP 7.1, the void return type was added. While PHP has no actual void value, it is generally understood across programming languages that a function that returns nothing is returning void. This should not be confused with returning null, as null is a value that can be returned.

function lacks_return(): void {
    // valid
}

Note that if you declare a void return, you cannot return any values or you will get a fatal error:

function should_return_nothing(): void {
    return null; // Fatal error: A void function must not return a value
}

However, using return to exit the function is valid:

function returns_nothing(): void {
    return; // valid
}

Syntax

• function f(ClassName $param) {}
• function f(bool $param) {}
• function f(int $param) {}
• function f(float $param) {}
• function f(string $param) {}
• function f(self $param) {}
• function f(callable $param) {}
• function f(array $param) {}
• function f(?type_name $param) {}
• function f() : type_name {}
• function f() : void {}
• function f() : ?type_name {}

Remarks

Type hinting or type declarations (opens new window) are a defensive programming practice that ensures a function's parameters are of a specified type. This is particularly useful when type hinting for an interface because it allows the function to guarantee that a provided parameter will have the same methods as are required in the interface.

Passing the incorrect type to a type hinted function will lead to a fatal error:

Fatal error: Uncaught TypeError: Argument **X** passed to **foo()** must be of the type **RequiredType**, **ProvidedType** given