Type hinting for any object

Question

I\'ve been working on code that\'s intended to be used with objects, without really caring what the kind of object is. I wanted to type hint that the method being written ex

Answer 1

As of php 7.2 this feature has now been implemented. you can type hint for any object now.

function myFunc(Object $myObject) : Object {
    return $myObject;
}

You can review this in the official documentation

Answer 2

stdClass is NOT a base class! PHP classes do not automatically inherit from any class. All classes are standalone, unless they explicitly extend another class. PHP differs from many object-oriented languages in this respect.

Answer 3

Although there is no type hinting for objects, you can use:

if (!is_object($arg)) {
    return;
}

Answer 4

There is no base class that all objects extend from. You should just remove the typehint and document the expected type in the @param annotation.

Answer 5

You could do something like this:

function myFunc ($obj)
{
     if ($obj instanceof stdClass) { .... }
}

Answer 6

Well it only took eight years, but this will soon be possible: PHP 7.2 introduces the object type hint! As I write this, it's currently in the RFC stage, and is due to be released in November.

Update, 30th November: PHP 7.2 has been released

RFC: Object typehint

Discussion

This behaves exactly as you might expect:

<?php

class Foo {}
class Bar {}

function takeObject(object $obj) {
    var_dump(get_class($obj));
}

takeObject(new Foo);
takeObject(new Bar);
takeObject('not an object');

Will result in:

string(3) "Foo"

string(3) "Bar"

Fatal error: Uncaught TypeError: Argument 1 passed to takeObject() must be an object, string given, called in...

See https://3v4l.org/Svuij

One side-effect of this is that object is now a reserved word, which unfortunately renders @Gaz_Edge's existing solution above broken. Fortunately, all you have to do to fix it is delete the interface.