Difference Between Base b = new Base(); AND Base b = new Derived(); ???

Question

I want to clarify the following, please help!!!
class A
{ public void f()
{ System.out.println("A->f");
class B extends A
{ public void g()
{ System.out.println("B->g");
public class Test
{ public static void main(String[] args)
{ A a = new A();
     A a = new B();
a.f();
//a.g();
A a = new A(); //(1)
means create an instance of class A of type A. Correct??
A b = new B(); //(2)
means create an instance of class B of type A
In both (1) and (2), instance "a" can access any methods of class A, but
cannot access any methods of class B
What's the difference then???

Answer 1

I want to clarify the following, please help!!!
class A
{ public void f()
{ System.out.println("A->f");
class B extends A
{ public void g()
{ System.out.println("B->g");
public class Test
{ public static void main(String[] args)
{ A a = new A();
     A a = new B();
a.f();
//a.g();
A a = new A(); //(1)
means create an instance of class A of type A.
Correct??Not quite. It means "define a variable 'a' of type 'class A', create an instance of 'class A' and assign the reference to that instance to variable 'a'.
A b = new B(); //(2)
means create an instance of class B of type ANot quite. It means "define a variable 'b' of type 'class A', create an instance of 'class B' and assign the reference to that instance to variable 'b'.
In both (1) and (2), instance "a" can access any
methods of class A, but
cannot access any methods of class BWell, the object actually referenced by the variable of type 'class A' is really an instance of class B and so if class B implemented a different behavior for any method defined in class A, then calling that method through a variable of class A would still get the behavior of class B - i.e. you get polymorphism.
Chuck