Java: extending a class and implementing an interface that have the same method

Question

Probably the following cannot be done (I am getting a compilation error: \"The inherited method A.doSomthing(int) cannot hide the public abstract method in B\"):

Answer 1

The method doSomethis() is package-private in class A:

public class A {
    int doSomthing(int x) { // this is package-private
        return x;
    }
}

But it is public in the interface B:

public interface B {
    int doSomthing(int x); // this here is public by default
}

Compiler is taking the doSomething() inherited by C from A which is package-private as the implementation of the one in B which is public. That's why it's complaining -

"The inherited method A.doSomthing(int) cannot hide the public abstract method in B"

Because, while overriding a method you can not narrow down the access level of the method.

Solution is easy, in class C -

@Override
public int doSomthing(int x) {
    // ...
}

Answer 2

make the method public

public class C extends A implements B {

    //trying to override doSomthing...

    public int myMethod(int x) {
        return doSomthingElse(x);
    }
}

interface methods are always public

or just use composition instead of inheritance

Answer 3

This has to do with visibility. You are using default (no modifier) visibility in C for myMethod but it needs to be public according to the interface B.

Now you might think you used the default visibility for all of them, since in neither A, B, nor C did you explicitly select one of public, private, or protected. However, the interface uses public whether or not you explicitly indicate so.

Answer 4

Simply making the method public when overriding it in C will do.