Inline interface implementation - Implement interface methods at declaration

Question

I come from java, where we can do something like this:

Action.java:

public interface Action {
    public void performAction();
}

MainClass.java:

public class MainClass {
    public static void main(String[] args) { //program entry point
        Action action = new Action() {

            public void performAction() {
                // custom implementation of the performAction method
            }

        };

        action.performAction(); //will execute the implemented method
    }
}

As you can see, I'm not creating a class which implements Action, but I'm implementing the interface directly on declaration.

Is something like this even possible with PHP?

What I've tried:

action.php:

<?php

interface Action {

    public function performAction();
}

?>

myactions.php:

include "action.php";

$action = new Action() {

    public function performAction() {
        //do some stuff
    }
};

What I get:

Parse error: syntax error, unexpected '{' in myactions.php on line 3

So, my question is: is something like this possible with PHP? How should I do it?

demonkoryu · Accepted Answer · 2018-09-14 14:13:30Z

5

With PHP 7, this has become possible with anonymous classes.

$action = new class implements Action() {
    public function performAction() {
        //do some stuff
    }
};

edited Sep 14, 2018 at 14:13

answered Dec 29, 2017 at 12:33

demonkoryu

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Comments

motanelu · Accepted Answer · 2014-10-31 21:55:00Z

No, can't. PHP doesn't offer anonymous classes like Java does. You can however try to simulate the behaviour you want, but the results will be...mixed at best.

Here's some code:

interface Action
{
    public function performAction();
}

class MyClass
{
    public function methodOne($object)
    {
        $object->performAction(); // can't call directly - fatal error

        // work around
        $closure = $object->performAction;
        $closure();
    }

    public function methodTwo(Action $object)
    {
        $object->performAction();
    }
}

$action = new stdClass();
$action->performAction = function() {
    echo 'Hello';
};

$test = new MyClass();
$test->methodOne($action); // will work
$test->methodTwo($action); // fatal error - parameter fails type hinting

var_dump(method_exists($action, 'performAction')); // false
var_dump(is_callable(array($action, 'performAction'))); // false

Hope it helps!

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