Type hinting for any object

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.

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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.

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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.

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

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

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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.

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There is no built-in mechanism to do this without requiring all users of your interface to extend a specified class. But why would you want to do this anyway? What do all object types have in common that's enough to make them suitable input for your API?

In all probability you wouldn't gain anything even if able to type hint like this. On the other hand, type hinting a parameter to implement an interface (such as Traversable) would be much more meaningful.

If you still want something akin to type hinting, the best you can do is substitute a runtime check with is_object on the parameter.

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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

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Typehint for stdClass works since PHP 5.3+ (if I am not wrong). Following is valid code using typehint for stdClass construct:

Example test.php:

class Test{
    function hello(stdClass $o){
        echo $o->name;
    }
}

class Arg2 extends stdClass{
    public $name = 'John';
    function sayHello(){
        echo 'Hello world!';
    }
}

$Arg1 = new stdClass();
$Arg1->name = 'Peter';

$Arg2 = new Arg2();
$Arg2->sayHello();

$test = new Test();

// OK
$test->hello($Arg1);
$test->hello($Arg2);

// fails
$test->hello(1);

Prints out:

Hello world! Peter John Catchable fatal error: Argument 1 passed to Test::hello() must be an instance of stdClass, integer given, called in test.php on line 32 and defined in test.php on line 5

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You could do something like this:

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

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