Java 8 lambda Void argument

Question

Let\'s say I have the following functional interface in Java 8:

interface Action {
   U execute(T t);
}

And for some cases I ne

Answer 1

That is not possible. A function that has a non-void return type (even if it's Void) has to return a value. However you could add static methods to Action that allows you to "create" a Action:

interface Action<T, U> {
   U execute(T t);

   public static Action<Void, Void> create(Runnable r) {
       return (t) -> {r.run(); return null;};
   }

   public static <T, U> Action<T, U> create(Action<T, U> action) {
       return action;
   } 
}

That would allow you to write the following:

// create action from Runnable
Action.create(()-> System.out.println("Hello World")).execute(null);
// create normal action
System.out.println(Action.create((Integer i) -> "number: " + i).execute(100));

Answer 2

Add a static method inside your functional interface

package example;

interface Action<T, U> {
       U execute(T t);
       static  Action<Void,Void> invoke(Runnable runnable){
           return (v) -> {
               runnable.run();
                return null;
            };         
       }
    }

public class Lambda {


    public static void main(String[] args) {

        Action<Void, Void> a = Action.invoke(() -> System.out.println("Do nothing!"));
        Void t = null;
        a.execute(t);
    }

}

Output

Do nothing!

Answer 3

The syntax you're after is possible with a little helper function that converts a Runnable into Action<Void, Void> (you can place it in Action for example):

public static Action<Void, Void> action(Runnable runnable) {
    return (v) -> {
        runnable.run();
        return null;
    };
}

// Somewhere else in your code
 Action<Void, Void> action = action(() -> System.out.println("foo"));

Answer 4

I don't think it is possible, because function definitions do not match in your example.

Your lambda expression is evaluated exactly as

void action() { }

whereas your declaration looks like

Void action(Void v) {
    //must return Void type.
}

as an example, if you have following interface

public interface VoidInterface {
    public Void action(Void v);
}

the only kind of function (while instantiating) that will be compatibile looks like

new VoidInterface() {
    public Void action(Void v) {
        //do something
        return v;
    }
}

and either lack of return statement or argument will give you a compiler error.

Therefore, if you declare a function which takes an argument and returns one, I think it is impossible to convert it to function which does neither of mentioned above.

Answer 5

The lambda:

() -> { System.out.println("Do nothing!"); };

actually represents an implementation for an interface like:

public interface Something {
    void action();
}

which is completely different than the one you've defined. That's why you get an error.

Since you can't extend your @FunctionalInterface, nor introduce a brand new one, then I think you don't have much options. You can use the Optional<T> interfaces to denote that some of the values (return type or method parameter) is missing, though. However, this won't make the lambda body simpler.

Answer 6

You can create a sub-interface for that special case:

interface Command extends Action<Void, Void> {
  default Void execute(Void v) {
    execute();
    return null;
  }
  void execute();
}

It uses a default method to override the inherited parameterized method Void execute(Void), delegating the call to the simpler method void execute().

The result is that it's much simpler to use:

Command c = () -> System.out.println("Do nothing!");