Question

Is there anything to stop me from type hinting like this within the __construct() parentheses?

<?php

    class SomeClass extends BaseClass {

        public function __construct(array $someArray) {

            parent::__construct($someArray);
        }

Or can I only do it like this?

<?php

    class SomeClass extends BaseClass {

        public function __construct($someArray = array()) {

            parent::__construct($someArray);
        }

Edit:

This is what works: (Thanks @hakra and @Leigh)

<?php

    class SomeClass extends BaseClass {

        public function __construct( array $someArray = NULL ) {

            parent::__construct( (array) $someArray);
        }

To me it looks nice and clean and I know exactly what it supposed to mean.

Answer 1

You can do so:

public function __construct(array $someArray = NULL)
{
    $someArray = (array) $someArray;
    parent::__construct($someArray);
    ...

The case here is to make use of the = NULL exception of the rule. The next line:

    $someArray = (array) $someArray;

is just some shorthand to convert NULL into an empty array and otherwise leave the array an array as-is (a so called cast^Docs to array):

Converting NULL to an array results in an empty array.

(from: Converting to array^Docs)

Sure you could write it even shorter:

public function __construct(array $someArray = NULL)
{
    parent::__construct((array) $someArray);
}

but it does not explain it that well.