PHP: Return value declaration in namespaces

Question

I recently started learning PHP and since I'm used to strongly typed languages (C++, Java) I would really prefer to make use of type hinting and return type declarations. However when I'm trying to use return type declaration for primitives I get a fatal error, stating that my function does not return the correct type.

Here is a code example:

<?php
namespace Foo;

class Bar {

    public function bar() : boolean {
        return true;
    }

}

$bar = new Bar();
$bar->bar();

?>

This will cause a fatal error, stating that Bar::bar does not return a value of Foo\boolean type. So I tried resorting to the global namespace by changing the function prototype to:

public function bar() : \boolean

But then I got a fatal error stating that my function does not return a value of type boolean. I guess PHP is looking for a type called boolean in the global namespace and not for the primitive type. Great, so how am I supposed to indicate that I want my function to return a value of the primitive type boolean?

You should also put declare(strict_types=1); as the first line of each file. — Richard Turner
– Richard Turner, Commented Feb 19, 2017 at 10:59

Richard Turner · Accepted Answer · 2017-02-19 10:53:29Z

8

The type hint for a Boolean value is actually bool. Try this:

<?php
namespace Foo;

class Bar {
    public function bar() : bool {
        return true;
    }
}

$bar = new Bar();
$bar->bar();

answered Feb 19, 2017 at 10:53

Richard Turner

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2 Comments

Krisztián Szabó Over a year ago

Damn, I feel stupid. Thank you for the clarification :)

Richard Turner Over a year ago

Nah, it's not stupid :) PHP allows classes to start with a lower-case letter, and until it started checking scalar type hints it was very common to see bool type-hinted as boolean (in comments). This is another case of a gotcha where the tooling can do little to help you spot an error.

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