Type Hinting: Default Parameters

Question

You can’t typehint strings, you can only typehint objects and arrays, so this is incorrect:

function setName ( string $name = "happ") {

(The reason you don’t get a compile-time error here is because PHP is interpreting “string” as the name of a class.)

The wording in the docs means that if you do this:

function foo(Foo $arg) {

Then the argument passed to foo() must be an instance of object Foo. But if you do this:

function foo(Foo $arg = null) {

Then the argument passed to foo() can either be an instance of object Foo, or null. Note also that if you do this:

function foo(array $foo = array(1, 2, 3))

Then you can’t call foo(null). If you want this functionality, you can do something like this:

function foo(array $foo = null) {
    if ($foo === null) {
        $foo = array(1, 2, 3);
    }

[Edit 1] As of PHP 5.4, you can typehint callable:

function foo(callable $callback) {
    call_user_func($callback);
}

[Edit 2] As of PHP 7.0, you can typehint bool, float, int, and string. This makes the code in the question valid syntax. As of PHP 7.1, you can typehint iterable.

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