This appears to create an object from an interface; how does it work?

Question

interface Int {
    public void show();
}

public class Test {     
    public static void main(String[] args) {
        Int t1 = new Int() {
            public void         


        

Answer 1

What this special syntax for anonymous inner classes does under the hood is create a class called Test$1. You can find that class file in your class folder next to the Test class, and if you printed t1.getClass().getName() you could also see that.

Answer 2

i think your object has nothing to do with the interface. If you comment out the whole interface, still you will get the same output. Its just anonymous class created. I think, unless you use the class "implements" you cant implement the interface. But i dunno how naming collision doesn't happen in your case.

Answer 3

You're defining an anonymous class that implements the interface Int, and immediately creating an object of type thatAnonymousClassYouJustMade.

Answer 4

This notation is shorthand for

Int t1 = new MyIntClass();

// Plus this class declaration added to class Test
private static class MyIntClass implements Int
    public void show() {
        System.out.println("message");
    }
}

So in the end you're creating an instance of a concrete class, whose behavior you defined inline.

You can do this with abstract classes too, by providing implementations for all the abstract methods inline.