In interface ONE I have a method A and in interface TWO I have method B. Both the methods are implemented in class Three. Now I assign an instance of Three to ONE, but still can I call method B of SECOND?
Even if this is possbile, is it correct?
Assuming you have this:
public interface A
{
public void methodA();
}
public interface B
{
public void methodB();
}
public class C implements A,B
{
public void methodA(){...}
public void methodB(){...}
}
You should be able to do this:
A a = new C();
a.methodA();
but not this:
a.methodB()
On the other hand, you can do this:
B b = new C();
b.methodB();
but not this:
b.methodA();
EDIT:
This is because you define object a as of being an instance of A. Although you are using a concrete class for the initialization (new C()), you are programming to an interface so only the methods defined in that interface will be visible.
Also if One extends Two, you will be able to do that. This may not be a solution, but I am only pointing to another way you can do that.
interface Two
{
void a();
}
interface One extends Two
{
void b();
}
class Three implements One
{
@Override
public void b() {}
@Override
public void a() {}
}
Then you can have
One one = new Three();
one.a();
one.b();
Remember that you're only able to call methods that are available to the assigned class/interface - it doesnt matter what methods the actual object supports, as far as the compiler is concerned it only looks at the assigned reference and what methods it has.
So, in your case if you assign:
Three three = new Three(); // all of the methods in One, Two and Three (if any) can be invoked here
One one = three; // Only the methods on One can be invoked here
Two two = three; // Only the methods on Two can be invoked here
