Question

public class B {

    static int i =1;

    public static int multiply(int a,int b)
    {
        return i;
    }

    public int multiply1(int a,int b)
    {
        return i;
    }

    public static void main(String args[])
    {
        B b = new A();
        System.out.println(b.multiply(5,2));
        System.out.println(b.multiply1(5,2));
    }
}

class A extends B
{
    static int i =8;

    public static int multiply(int a,int b)
    {
        return 5*i;
    }

    public int multiply1(int a,int b)
    {
        return 5*i;
    }

}

Output:

1

40

Why is it so? Please explain.

Answer 1

Calling static methods via references is a really bad idea – it makes the code confusing.

These lines:

B b = new A();
System.out.println(b.multiply(5,2));
System.out.println(b.multiply1(5,2));

are compiled into this:

B b = new A();
System.out.println(B.multiply(5,2));
System.out.println(B.multiply1(5,2));

Note that the calls don’t rely on b at all. In fact, b could be null. The binding is performed based on the compile-time type of b, ignoring its value at execution time.

Polymorphism simply doesn’t happen with static methods.