q from jiris.com on overriding

[JayaSiji Gopal]

What is the output of trying to compile and run the following code?
(Select two correct answers)
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public class Test013 extends Super
{
void method() { System.out.println("Here"); }
public static void main(String args[])
{
Super s = new Test013();
((Test013)s).method(); //1
}
}
class Super{ }
--------------------------------------------------------------------------------
A: The code does not compile because of line //1.
B: The code can be made compilable by
removing the cast in line //1.
C: The code compiles and runs with output: Here.
D: The cast in line //1 is not necessary.
E: The cast in line //1 is needed.

I answered c and d, but the correct answer is c and e.
In line 1, the variable s at compile time is of reference type Super and at runtime is of type Test013. All method invocations are resolved at runtime. Hence, I feel the cast is not necessary.
Am i right? any mistakes?

[Animesh Shrivastava]

Hello,
As u only said that during compile time the variable s is of reference type Super. So, while compilation it checks whether the method method() is present or not for the class Super.
And therefore it gives a compilation error.
So, i guess the correct answers are C and E

[Adeeb Abdul Karim]

Test013 object is created and a variable of type Super is holding it. This happens at runtime.

At compiler time the method is not visible for a variable of type Super, because it is not defined in it. So if we want to use the method, we have to cast it to a type where the method is defined, then only the compiler will realise that it is having such a method.

So E is correct.

[Barry Gaunt]

Yes, it's a simple as that: class Super does not have a method called "method".