CODEWITHSUNDEEP
javalambdajava-8functional-programmingfunctional-interface
vinod827
vinod827

Reputation: 1544

Unable to override a default method of an Interface using Lambda expression

I have a functional Interface which contains 1 static method, 1 default method and 1 non-static abstract public method like below :-

package com.lambda;

@FunctionalInterface
public interface MyFunctionalInterface {

    static void printDetails(){
       System.out.println("Im a static method");
    }

    default void printDetails2(){
        System.out.println("I'm in default method");
    }

    void printName();

}

From a different class, using Lambda expression, I'm trying to invoke and override these methods like below :-

package com.lambda;

public class Example1 {

    public static void main(String[] args) {
        //static method call
        MyFunctionalInterface.printDetails();

        //Lambda expression call to an abstract method
        MyFunctionalInterface myFunctionalInterface = () -> System.out.println("My name is Vinod Kumar Nair");
        myFunctionalInterface.printName();

        //Lambda expression call to a default method
        MyFunctionalInterface myFunctionalInterface1 = () -> System.out.println("Im overriding default method");
        myFunctionalInterface1.printDetails2();



    }
}

Here is the output, I'm getting :-

Im a static method

My name is Vinod Kumar Nair

I'm in default method

Using Lambda expression, I can invoke the static method called 'printDetails'. I can also override the 'printName()' with my implementation logic, where 'My name is Vinod Kumar Nair' is getting printed.

The problem is I'm not able to override the default method, printDetails2() with my implementation. I'm not getting the output as 'Im overriding default method'. I'm still getting the default implementation logic.

Is there any issue in overriding the default method using Lambda expression. If yes, then you can please tell me how to override this default method using the Lambda expression.

Thank you

Upvotes: 2

Views: 3411

Answers (2)

Alexey Romanov
Alexey Romanov

Reputation: 170745

You can't. Lambda expressions in Java can only be used for an interface with a single abstract method (in this case printName) and will be used as the implementation of this abstract method. Note that any instance of MyFunctionalInterface needs to implement printName; where do you want it to come from if your lambda was used for printDetails2?

If you want to implement any default methods, you have to use an anonymous class:

MyFunctionalInterface myFunctionalInterface1 = new MyFunctionalInterface {
    @Override 
    void printDetails2() { ... }

    @Override 
    void printName() { ... }
};

("have to" in this context: when you want to create it in place and assign to a variable.)

Upvotes: 4

dehasi
dehasi

Reputation: 2773

@FunctionalInterface means that you have only one method to implement.

When you do 

 MyFunctionalInterface myFunctionalInterface1 = () -> System.out.println("Im 
overriding default method");

You implement the method, but not override any default method.

To override default methods you have to create a class which implements your interface and overrides the method there.

public class MyClass implements MyFunctionalInterface {
    @Override 
    void printDetails2() { ... }

    @Override 
    void printName() { ... }
}

Upvotes: 4

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