java - How does adding an unused generic parameter to an interface stop a class implementing it? -

the following code

public interface igiveup {     void surrender(list<class> l); } public class giveup implements igiveup {      @override public void surrender(list<class> l) {} } 

compiles fine. when add unused generic type parameter interface

public interface igiveup<x> {     void surrender(list<class> l); } 

it fails compile (javac 1.6.0_23)

igiveup.giveup not abstract , not override abstract method surrender(java.util.list) 

it compile if either specify generic in implementation

public class giveup implements igiveup<object> 

or make method parameter list of not generic type

void surrender(list l); 

your class trying implement raw type igiveup - raw type doesn't know generics, method signature after type erasure just:

void surrender(list l) 

it doesn't matter method parameter didn't use type parameter interface declaration: type erasure removes all traces of generics signatures.

basically, should avoid raw types far possible. more details, follow links above sections of jls, or read java generics faq.