Extending an interface

Question

I downloaded a code that has a function that starts like this:

public class MDP<S, A extends Action> implements MarkovDecisionProcess<S, A> {
//some code...blah blah blah...
}

S and A are supposed to be some kind of type. Action is an interface.

In my code, I want to use the given class MDP. Therefore, I had to define S and A; I defined S to be a certain class, but I don't know how to define A... it isn't a class and it isn't an interface. What's it supposed to be?

Thanks :)

Answer 1

"A" type would be a class that implements an interface which extends Action.

Answer 2

For your code

public class MDP<S, A extends Action> implements MarkovDecisionProcess<S, A> {
  //some code...blah blah blah...
}

S and A are types. They can refer to an interface or a Class. The letter designations are arbitrary. They can be any letter. This is just as valid

public class MDP<Q, Z extends Action> implements MarkovDecisionProcess<Q, Z> {
  //some code...blah blah blah...
}

All it says is, "class MDP is a parameterized class, having the parameters S and A where A is some subclass of Action, and it implements the interface MarkovDecisionProcess<S,A>. S and A may or may not be of the same type."

When you give type parameters to the class, you can narrow the specificity of the type. In this case, the S stays the same (in terms of specificity) but you narrow second parameter to some instance of Action.

It could be possible to have those types on the interface declaration. They could then be left off on the instance.

public interface MarkovDecisionProcess<S,V extends Action>{}

public class MDP implements MarkovDecisionProcess{}

You might benefit from reading some of the official documentation.