Overriding member variables in Java ( Variable Hiding)

Question

I am studying overriding member functions in JAVA and thought about experimenting with overriding member variables.

So, I defined classes

public clas         


        

Answer 1

From JLS Java SE 7 Edition §15.11.1:

This lack of dynamic lookup for field accesses allows programs to be run efficiently with straightforward implementations. The power of late binding and overriding is available, but only when instance methods are used.

Answers from Oliver Charlesworth and Marko Topolnik are correct, I would like to elaborate a little bit more on the why part of the question:

In Java class members are accessed according the type of the reference and not the type of the actual object. For the same reason, if you had a someOtherMethodInB() in class B, you wouldn't be able to access it from aRef after aRef = b is run. Identifiers (ie class, variable, etc names) are resolved at compile time and thus the compiler relies on the reference type to do this.

Now in your example, when running System.out.println(aRef.intVal); it prints the value of intVal defined in A because this is the type of the reference you use to access it. The compiler sees that aRef is of type A and that's the intVal it will access. Don't forget that you have both fields in the instances of B. JLS also has an example similar to yours, "15.11.1-1. Static Binding for Field Access" if you want to take a look.

But why do methods behave differently? The answer is that for methods, Java uses late binding. That means that at compile time, it finds the most suitable method to search for during the runtime. The search involves the case of the method being overridden in some class.