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I have a class defined by an interface

public interface Test {
    void testMethod();
}

Test test = new TestImpl();

public class TestImpl implements Test {
    @Override
    public void testMethod() {
         //Nothing to do here
    }

    public void anotherMethod() {
        //I am adding this method in the implementation only.
    }
}

How can I call anotherMethod?

test.anotherMethod(); //Does not work.

I want to be able to define a few methods in the implementation only because in my production code, the Test interface covers a pretty broad spectrum of classes and is implemented by multiple classes. I use methods defined in the implementation to set dependencies that aren't covered by the DI framework in my unit testing so the methods change from implementation to implementation.

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1 Answer

The problem is with the following line:

Test test = new TestImpl();

This tells the compiler to forget that the new object is a TestImpl and treat it as a plain old Test. As you know, Test does not have anotherMethod().

What you did is called "upcasting" (casting an object to a more general type). As another poster has said, you can fix your problem by not upcasting:

TestImpl test = new TestImpl();

If you're sure that a Test object is really a TestImpl, you can downcast it (tell the compiler it is a more specific type):

Test test = new TestImpl();
:
((TestImpl) test).anotherMethod();

This is generally a bad idea, however, since it can cause ClassCastException. Work with the compiler, not against it.


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