Question

class Base 
{
    int i = 99;
    public void amethod()
    {
        System.out.println("Base.amethod()");
    }

    Base()
    {
        amethod();
    }
}
public class Derived extends Base
{
    int i = -1;

    public static void main(String argv[])
    {
        Base b = new Derived();
        System.out.println(b.i);
        b.amethod();
    }

    public void amethod()
    {
        System.out.println("Derived.amethod()");
    }
}

Why does this code print b.i = 99 and not b.i = -1?
Thank you.

Answer 1

It’s because you are referencing a member variable of the object instead of a method. Since it is declared as a Base object, it will use Base.i because member variables do not benefit from polymorphism that is afforded to methods. If you added a getI() method to both classes and called that instead of just b.i, it would work as you expect.

public int getI() {
    return this.i;
}


    Base b = new Derived();
    System.out.println(b.getI());