Why do I get the value of the default method and not the overridden one?

Question

interface MyInterface {
    default int someMethod() {
        return 0;
    }

    int anotherMethod();
}

class Test implements MyInterface {
    public static void main(String[] args) {
        Test q = new Test();
        q.run();
    }

    @Override
    public int anotherMethod() {
        return 1;
    }
    
    void run() {
        MyInterface a = () -> someMethod();
        System.out.println(a.anotherMethod());
    }
}

The execution result will be 0, although I expected 1. I don't understand why the result of the overridden method is not returned, but the result of the default method is returned.

Answer 1

Becouse you used another implementation with lambda, where implementation of another method is "some method". You can create interfaces with lambda. On this case you do

MyInterface a = new Test(); System.out.println(a.anotherMethod());

or :

    MyInterface a = () -> 1;
    System.out.println(a.anotherMethod());

Answer 2

The statement MyInterface a = () -> someMethod(); is more or less equivalent to the following code:

MyInterface a = new MyInterface() {
  @Override int anotherMethod() {
    return someMethod();
  }
};

someMethod() is not overridden anywhere in your example, so the default implementation is used. The default implementation of return 0. So calling a.anotherMethod() is expected to return 0.

The overridden Test#anotherMethod is never called by your code. If you want to execute this method, you must call this.anotherMethod() (or assign MyInterface a = this;).

Answer 3

Because you didn't call the overridden method.

MyInterface a = () -> someMethod();
System.out.println(a.anotherMethod());

a has nothing to do with the Test instance in which it is called.