How to check if an interface exists in PHP without creating an instance?

Question

PHP

The class name comes to the method, it is necessary to check this class for the implementation of the interface with minimal effort. Right now I'm doing (new $class) instanceof $interface . But it seems to me a bit unreasonable to create an instance of a class just to test the interface. Moreover, I do not want to create a Reflection object because it looks even worse. Any ideas how to make it less time consuming for the system?

Answer 1

interface MyInterface { }
class MyClass implements MyInterface { }
$interfaces = class_implements('MyClass');
if($interfaces && in_array('MyInterface', $interfaces)) {
// Class MyClass implements interface MyInterface
}
Thanks adamsafr for the answer :

Fits?
stackoverflow.com/questions/20169805/php-check-if-...