Develop Reference

Object class inheritance

We know that class A extends Object class implicitly but if we instantiate Object class object with class A constructor and try to access print() method of class A it shows a compile time error while in other case where class B extends class A and Class A object instantiated with class B constructor and try to access print() method works fine.
class A{
public void print(){
System.out.println("A");
}
}
class B extends A{
public void print(){
System.out.println("B");
}
}
public class MyClass {
public static void main(String args[]) {
Object o = new A();
o.print(); // error
A a = new B();
a.print(); // print B
}
}

This line:
o.print(); // error
You know that o is actually of type A.
The compiler could know, but in Java, it does not know. The compiler only knows that there is some variable named o, of type Object.
Thus the compiler slaps your fingers for invoking a method on o that the declared type Object doesn't support!
Other languages are "smarter" about such things, but in Java, that is how things are.

You example with class B extends A isn't the same as the case of A extends Object. The difference is that in the first case the superclass (A) contains the print() method, in the second case the superclass (Object) doesn't contain the print(). If you remove the print() method from the A class, the second part won't work either:
class A{
}
class B extends A{
public void print(){
System.out.println("B");
}
}
public class MyClass {
public static void main(String args[]) {
Object o = new A();
//o.print(); // error
A a = new B();
a.print(); // error
}
}

Because Object class does not have print() method and the reference o is of type Object, the compiler will throw an error.
In case 2 : class A has print() method thats why no compilation error and at runtime it calls class B's print() method because we assigned class B's reference at runtime.

what's the use of java virtual method invokation?

Say I have the following code
public class A {
public void callme() {
System.out.println("Calling of class A function ");
}
}
public class B extends A {
public void callme() {
System.out.println(" Calling of class B fuction ");
}
public void Hello() {
System.out.println("hello guys");
}
}
and a main() that does the following
public class Main {
public static void main(String [] args){
A a = new B();
B b = new B();
b = (B)a;
a.callme();
b.callme();
a.Hello(); // show error : Hello is undefined for method A
}
}
The book says "you get the behavior associated with the object to which the variable refers at runtime". Ok, I get behavior of B class when a method callme is called, but when I access the method Hello, it shows an error that Hello is undefined for method A. why is that?

Polymorphism doesn't work this way. Since A is parent of B, B can inherit methods of A (like son can inherit properties of father), but its not vice versa because A doesn't know which classes are inheriting it (A does not know who are its children.)
For Example, suppose there is one more class C:
public class C extends A {
public void callme(){
System.out.println(" Calling of class C fuction ");
}
public void Hello(){
System.out.println("hello guys, I am C");
}
}
Now, if you use
a.Hello();
how will a know which child class it should call since it does not methods of its child. only its own abstract method, which it knows child will implement for sure.

a is of type A, even if it is referring to an instance of B.
A doesn't have the Hello method (even an abstract declaration). So your compiler emits the error. Adjusting your code slightly to
public abstract class A {
public abstract void Hello();
/*the rest as before*/
would be a fix. A is then assuming the characteristics of an interface.

public class B extends A {
public void callme(){
System.out.println(" Calling of class B fuction ");
}
public void Hello(){ // hello method is only the part of class B not A.
System.out.println("hello guys");
}
}
In above class hello() method is a part of B only. it not override by the method of A.
now in your main method call
public static void main(String [] args){
A a= new B(); // object of b referred by (a) Reference Variable of A
B b= new B(); // object of b referred by (b) Reference Variable of B
b= (B)a;
a.callme(); //override takes place and method of B's Callme() called
b.callme(); //again override takes place here and method of B's Callme() called
a.Hello();// buttttttt
b.Hello(); // this will compile and executes fine.
}
}
Here you used the reference variable of class A which don't have any method name Hello(). So, Method Resolution will not take place(won't be able to find any method like Hello()).
but if you try to call b.Hello() using the reference variable of B then it will work fine to you.
Now suppose there is another class C which is a sub class of A and contains a Method Name Hello().
public class C extends A {
public void Hello(){ // hello method is only the part of class B not A.
System.out.println("hello guys");
}
}
In main a Statement like this
A a = new C();
and if you try to call a.Hello() then which Hello() method will call. Compiler will get confused.
So, This concept work only when you try to override the method of super class in sub Class.

Is the parent class aware of classes derived from it ?
Casting does not change the actual object type. Only the reference type gets changed.
I highly recommend you my upcasting and downcasting writing from the link

If a class inherits a method does calling that method on that class point to the super class or does it make a local copy within the child class?

Say I have Class A and Class B. Class B extends Class A. Class A has one method.
public class notimportant
{
public void one()
{
}
}
public class A extends notimportant
{
public void one()
{
//assume there is a super class making this call legal which doesnt do anything
super.one();
System.out.println("blah");
}
}
public class B extends A
{
}
A var1 = new B();
if I call 'var1.one();' will the output end up being:
"blah"
"blah"
because it creates a local copy of 'one()' in Class B and then reads that which calls 'super()' which leads it up to method 'one()' in Class A OR does it just print
"blah"
because it knows to look directly at Class A
EDIT: Hope that is a lot more clear now.

It will follow the way you have it currently written:
-> New object of class B
-> Call method One on this object
-> First line calls supermethod, proceed to execute it
-> Second line prints out after that
Your code doesn't compile at all though, you might want to clear that up. What's keeping you from testing this yourself?
Here's the new situation as you described it. Everything still works as expected, you just add a layer.
public class C {
public void test() {
System.out.println("Inside C");
}
}
public class B extends C {
public void test() {
super.test();
System.out.println("Inside B");
}
}
public class A extends B {
public static void main(String[] args) {
A obj = new A();
obj.test();
}
}
Output:
Inside C
Inside B

super means your superclass – it's resolved at compile-time.
It does not mean the immediate parent class of whatever the runtime type of this is.

Can a private method in super class be overridden in the sub-class?

Can private methods be overridden in Java?
If no, then how does the following code work?
class Base{
private void func(){
System.out.println("In Base Class func method !!");
};
}
class Derived extends Base{
public void func(){ // Is this a Method Overriding..????
System.out.println("In Derived Class func method");
}
}
class InheritDemo{
public static void main(String [] args){
Derived d = new Derived();
d.func();
}
}

No, you are not overriding it. You can check by trying to mark it with #Override, or by trying to make a call to super.func();. Both won't work; they throw compiler errors.
Furthermore, check this out:
class Base {
private void func(){
System.out.println("In base func method");
};
public void func2() {
System.out.println("func2");
func();
}
}
class Derived extends Base {
public void func(){ // Is this an overriding method?
System.out.println("In Derived Class func method");
}
}
class InheritDemo {
public static void main(String [] args) {
Derived D = new Derived();
D.func2();
}
}
It will print:
func2
In base func method
When you change func() in Base to public, then it will be an override, and the output will change to:
func2
In Derived Class func method

No, a private method cannot be overridden because the subclass doesn't inherit its parent's private members. You have declared a new method for your subclass that has no relation to the superclass method. One way to look at it is to ask yourself whether it would be legal to write super.func() in the Derived class. There is no way an overriding method would be banned from accessing the method it is overriding, but this would precisely be the case here.

No, it is not. You can mark an override just to make sure like this:
#Override
public void func(){
System.out.println("In Derived Class func method");
}
And in this case it would be a compiler error.

You are not overriding. You cannot override private members, you are merely defining a new method in Derived. Derived has no knowledge Base's implementation of func() since its declared as private. You won't get a compiler error when you define func() in Derived but that is because Derived does not know Base has an implementation of func(). To be clear: it would be incorrect to say you are overriding Base's implementation of func().

In addition to the already correct answer, consider this:
public class Private {
static class A {
public void doStuff() {
System.out.println(getStuff());
}
private String getStuff() {
return "A";
}
}
static class B extends A {
public String getStuff() {
return "B";
}
}
public static void main(String[] args) {
A a = new A();
a.doStuff();
a = new B();
a.doStuff();
B b = new B();
b.doStuff();
}
}
This will print
A
A
A
although B "overrides" getStuff(). As implementation of doStuff() is fixed to calling A#getStuff(), no polymorphism will be triggered.

Nope because if you do something like Base b = new Derived(); you still won't be able to call b.func(). What you're doing is called "hiding".

Since the method is private it is not visible to the other classes.Hence the derived class does not inherit this method.
So this is not the case of overriding

Method hiding will be happening here instead of overriding. like what happens in case of static.

Actually,you are not overriding.Before Java5
an overridden method's return type must match with parent class's method.
But Java 5 introduced a new facility called covariant return type.You can override a method with the same signature but returns a subclass of the object returned. In another words, a method in a subclass can return an object whose type is a subclass of the type returned by the method with the same signature in the superclass.
you can follow this thread :Can overridden methods differ in return type?

The private member of the base class cannot be access by anyone outside of the class and cannot be overridden. The function in the derive class is an independent function that can be access by anywhere.
The code would run the function in the derived class

A private method can never be over ridden. It is always hidden.
In your example - Derived class has one parent class private method and has its own function func. Both are different, and the func is not over ridden. Its a separate independent function.
If you create a new function in parent class calling parent class function, the parent func will be called, if parent class reference is used as opposed in the case of method over ridding
Note : An object defines the members which it has, and a reference defines which it can access
// Method Over ridding case
class Base{
public void func(){
System.out.println("Parent class");
};
public void func1(){
func();
}
}
class Derived extends Base{
public void func(){
System.out.println("Derived class");
}
}
class InheritDemo{
public static void main(String [] args){
Derived d = new Derived();
d.func1(); // Prints Derived class
Base b = new Derived();
b.func1(); // Prints Derived class - no matter parent reference is calling,as there as method is overridden - Check func1() is in parent class, but id doesn't call parent class func() as the compiler finds a func() method over ridden in derived class
}
}
// Method Hidding case - Private and static methods case
class Base{
private void func(){
System.out.println("Parent class");
};
public void func1(){
func()
}
}
class Derived extends Base{
public void func(){ // Is this a Method Overriding..????
System.out.println("Derived class");
}
}
class InheritDemo{
public static void main(String [] args){
Derived d = new Derived();
d.func1(); // Prints Derived class
Base b = new Derived();
b.func1();
// Prints Parent class - the reason is we are using the parent class reference, so compiler is looking for func() and it founds that there is one private class method which is available and is not over ridden, so it will call it. Caution - this won't happen if called using derived class reference.
b.func();
// this prints the Derived class - the compiler is looking func(), as Derived class has only one func() that it is implementing, so it will call that function.
}
}

Read comments in the below code snippet to find the answer.
Sources:
Definition reference:
Credits for the source code example(reference) from the book - "OCA Oracle Certified Associate Java SE 8 Programmer Study Guide Exam 1Z0-808 Book" from 'Jeanne Boyarsky' and 'Scott Selikoff'.
public class Deer {
public Deer() { System.out.print("Deer"); }
public Deer(int age) { System.out.print("DeerAge"); }
private boolean hasHorns() { return false; }
public static void main(String[] args) {
Deer deer = new Reindeer(5);
System.out.println(","+deer.hasHorns());// false is printed.
}
}
class Reindeer extends Deer {
public Reindeer(int age) { System.out.print("Reindeer"); }
private boolean hasHorns() { return true; } // Overriding possible, but is of no use in the below context.
// Below code is added by me for illustration purpose
public static void main(String[] args) {
Deer deer = new Reindeer(5);
//Below line gives compilation error.
//System.out.println(","+deer.hasHorns());
}
}

How can I call the constructor of the grand parent class?

How can we call the constructor of grand parent.
eg: B inherits from A and C inherits from B.
I need to call the constructor of A in C. Is it possible without creating an instance of B?
If i need this how this can be implemented in Java.

You can't invoke the constructor of A directly in the constructor of C.
You can (and actually must) call it indirectly, 'though. Each constructor of B has to call a constructor of A (either explicitly or implicitly). Since every constructor of C needs to call one of the constructors of B you will always call one of As constructors.

use super() from C and from B to access A's constructor
class A {
public A() {
System.out.println("A");
}
}
class B extends A {
public B() {
super();
System.out.println("B");
}
}
class C extends B {
public C() {
super();
System.out.println("C");
}
}
public class Inheritance {
public static void main(String[] args) {
C c = new C();
}
}
Output :
A
B
C
Note:
If all are default constructor then no need to write super(); it will implicitly call it.
If there is parametrized constructor then super(parameter.. ) is needed

C will have to call B's constructor and in turn B will call A's constructor...
there is No other option to call Grand parent's constructor

An example to Joachim's answer:
class A {
A() { System.out.print("A()")};
A(Object o) { System.out.print("A(Object)")};
}
class B {
B() { super(); System.out.print("B()")};
B(Object o) { super(o); System.out.print("B(Object)")};
}
class C {
C() { super(); System.out.print("C()")};
C(Object o) { super(o); System.out.print("C(Object)")};
}
Now calling:
C c = new C();
will produce: A()B()C(), while calling:
C c = new C(new Object());
will produce: A(Object)B(Object)C(Object).
As you can see, by manipulating super() call you can actually call explicit parent constructor.

Actually, if you have class B extending class A, you can't avoid to call the constructor method for A when you create an instance of the class B. If this mechanism doesn't seems clear to you, I recommend this reading about Java's inheritance system: http://download.oracle.com/javase/tutorial/java/IandI/subclasses.html .
Here's an example:
class A {
public A() {
doSomethingA();
}
}
class B extends A {
public B() {
super(); // this is the call to the A constructor method
doSomethingB();
}
}
class C extends B {
public C() {
super(); // this is the call to the B constructor method
doSomethingC();
}
}
Creating a new object with new C() will involve, in order, the calling of A() - B() - C()