Java 8 lambda Void argument

Question

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

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

And for some cases I need an action without arguments or return type. So I write something like this:

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

However, it gives me compile error, I need to write it as

Action<Void, Void> a = (Void v) -> { System.out.println("Do nothing!"); return null;};

Which is ugly. Is there any way to get rid of the Void type parameter?

Answer 1

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 2

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.