About method

Question

Look at the code below and can you tell me why the code prints out:
SuperClass Constructor Executed
SubClass Constructor Executed
methodA in Superclass
methodB in Superclass//Why not prints out methodB in Subclass,since methodB in sub has been overriden
9
public class Superclass {
Superclass() {
System.out.println("SuperClass Constructor Executed");
private int methodB() {
System.out.println("methodB in Superclass");
return 9;
int methodA() {
System.out.println("methodA in Superclass");
return methodB();
class Subclass extends Superclass {
Subclass() {
System.out.println("SubClass Constructor Executed");
public static void main(String[] args) {
System.out.println(new Subclass().methodA());
protected int methodB() {
System.out.println("methodB in Subclass");
return 1;
}

Answer 1

hi,but can you tell me how java virtual machine
identify methodB() when methodA calls is the version
of Superclass,not of Subclass?I'm not sure I understand the question?
If methodB is private or "package", then at compile time the compiler knows which method is to be called. If the method is protected or public, then it can be overridden and the check must be made at runtime; It doesn't matter which class the method is declared in - it is called on an object, not on the class! (note that you get different behaviour with static methods, which you don't actually override, but hide, and do infact call on the class rather than an object).
And can you tell me whether there is an instance of
Superclass is created when an instance of Subclass is
declared and created?No. Just a single instance of Subclass is created, but to properly initialise that instance the code in the Superclass constructor must be executed.