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"):

public class A {     int doSomthing(int x) {         return x;     } }  public interface B {     int doSomthing(int x); }  public class C extends A implements B {      //trying to override doSomthing...      int doSomthing(int x) {         return doSomthingElse(x);     } } 

Assuming I am allowed to change neither A nor B, my question is can I somehow define C in such a way that it will inherit from both A and B (suppose that it is required for some framework that C will be both an instance of A and B).

Or if not, how would you work around this?

Thanks!

Answer 1

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 2

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) {     // ... }