not implementing all of the methods of interface. is it possible?

Is there any way to NOT implement all of the methods of an interface in an inheriting class?

The only way around this is to declare your class as abstract and leave it to a subclass to implement the missing methods. But ultimately, someone in the chain has to implement it to meet the interface contract. If you truly do not need a particular method, you can implement it and then either return or throw some variety of NotImplementedException, whichever is more appropriate in your case.

The Interface could also specify some methods as 'default' and provide the corresponding method implementation within the Interface definition (https://docs.oracle.com/javase/tutorial/java/IandI/defaultmethods.html). These 'default' methods need not be mentioned while implementing the Interface.

The point of an interface is to guarantee that an object will outwardly behave as the interface specifies that it will

If you don't implement all methods of your interface, than you destroy the entire purpose of an interface.

We can override all the interface methods in abstract parent class and in child class override those methods only which is required by that particular child class.

Interface

public interface MyInterface{
    void method1();
    void method2();
    void method3();
}

Abstract Parent class

public abstract class Parent implements MyInterface{
@Override
public void method1(){

}
@Override
public void method2(){

}
@Override
public void method3(){

}
}

In your Child classes

public class Child1 extends Parent{
    @Override
    public void method1(){

    }
}




public class Child2 extends Parent{
    @Override
    public void method2(){

    }
}

I asked myself the same question, and then learned about Adapters. It solved my problem, maybe it can solve yours. This explains it very well : https://blogs.oracle.com/CoreJavaTechTips/entry/listeners_vs_adapters