How can we create object of interface in java? [duplicate]

Question

How is this code working i m totally puzzled....

package com.servletpack.examples;

interface check {
    public void message();
}
public class Interface {
    public static void main(String[] args) {
        try {
            check t = new check() {//how????????????????
                public void message() {
                    System.out.println("Method defined in the interface");
                }
            };
            t.message();
        } catch (Exception ex) {
            System.out.println("" + ex.getMessage());
        }
    }
}

Answer 1

It is anonymous class. Your check class is an interface. Anonymous class defines an implementation of given interface on the fly. So it saves you from creating a seperate class for Interface's implementation. This approach is only useful when you know you will never require this implementation any where else in the code.

Hope this explanation helps !!

Answer 2

With that syntax, you create an anonymous class, which is perfectly legal.

Internally, anonymous classes are compiled to a class of their own, called EnclosingClass$n where the enclosing class' name precedes the $ sign. and n increases for each additional anonymous class. This means that the following class is being created:

class Interface$1 implements check {
     public void message() {
         System.out.println("Method defined in the interface");
     }
}

Then, the code in main compiles to internally use the newly-defined anonymous class:

check t = new Interface$1();
t.message();