Question

I would like to know how to create a contract with the caller for the Method parameter in the event the method has parameters itself. So that I use…

ClassA {
  String string_ = "HI";

  public static void subscribe(Object class, Method action) {
    action.invoke(class, string_);
  }
}

ClassB {

  ClassB() {
    ClassA.subscribe(this, this.getClass().getMethod("load", String.class));
  }

  public void load(String input) {
    if(input.equals("HI")) { 
      ... 
    }
  }
}

I would like to know how to ensure the Method passed as “action” takes String as a parameter (i.e. ensure Method action == load(String){})? Is there something like this available:

public static void subscribe(Object class, Method action(String.class)) {

I want to do it in the method signature of subscribe so that it is obvious to the calling class (ClassB) that it needs to be prepared to take an argument of specified type.

EDIT: Updated last code bit so not to appear as if Method was generic. Poor choice of using <> on my part to represent an example of what I was trying to convey.

Answer 1

There’s no way to do that in Java. The Method class is not generic, and there is no way for it to be so, because methods can take any number of parameters, and there is no way to make a class generic over a variable number of types.

Probably the best you can do is to declare your own type to use instead of Method:

public interface Action<T, P> {
    public void invoke(T target, P parameter);
}

Then:

public static <T> void subscribe(T obj, Action<T, String> action) {
    action.invoke(obj, string_);
}

ClassB() {
    ClassA.subscribe(this, new Action<ClassB, String>() {
        public void invoke(ClassB target, String parameter) {
            target.load(parameter);
        }
    });
}