Type hinting for any object
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 expected an object of any type, but ran into some difficulty.
I tried function myFunc (object $obj)
and function myFunc (stdClass $obj)
but both of these generated errors when I tried to pass objects in:
Catchable fatal error: Argument 1 passed to MyClass::MyFunc() must be an instance of object, instance of ObjectActualClass given
The same happened with stdClass
as well
What am I missing? I thought that all classes that didn't explicitly inherit from another class inherited from stdClass
, meaning that the base class of every class in PHP would be stdClass
. Is this not the case?
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.
The best way to enforce this would be to create a degenerate interface called Object
. A degenerate interface means it has no defined methods.
interface Object {
// leave blank
}
Then in your base classes, you can implement Object
.
class SomeBase implements Object {
// your implementation
}
You can now call your function as you wanted to
function myFunc (Object $obj);
myFunc($someBase);
If you pass any object which inherits from your Object
interface, this type hint will pass. If you pass in an array, int, string etc, the type hint will fail.
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.
Although there is no type hinting for objects, you can use:
if (!is_object($arg)) {
return;
}