Output of Java Program | Set 20 (Inheritance)

Prerequisite – Inheritance in Java

Predict the output of following Java Programs.
Program 1 :

class A
{
    public A(String s)
    {
        System.out.print("A");
    }
}

public class B extends A
{
    public B(String s)
    {
        System.out.print("B");
    }
    public static void main(String[] args)
    {
        new B("C");
        System.out.println(" ");
    }
}

Output: Compilation fails

prog.java:12: error: constructor A in class A cannot be applied to given types;
    {
    ^
  required: String
  found: no arguments
  reason: actual and formal argument lists differ in length
1 error

Explanation: The implied super() call in B’s constructor cannot be satisfied because there isn’t a no-arg constructor in A. A default, no-arg constructor is generated by the compiler only if the class has no constructor defined explicitly.For detail See – Constructors in Java

Program 2 :

class Clidder
{
    private final void flipper()
    {
        System.out.println("Clidder");
    }
}

public class Clidlet extends Clidder
{
    public final void flipper()
    {
        System.out.println("Clidlet");
    }
    public static void main(String[] args)
    {
        new Clidlet().flipper();
    }
}

Output:

Clidlet

Explanation: Although a final method cannot be overridden, in this case, the method is private, and therefore hidden. The effect is that a new, accessible, method flipper is created. Therefore, no polymorphism occurs in this example, the method invoked is simply that of the child class, and no error occurs.

Program 3 :

class Alpha
{
    static String s = " ";
    protected Alpha()
    {
        s += "alpha ";
    }
}
class SubAlpha extends Alpha
{
    private SubAlpha()
    {
        s += "sub ";
    }
}

public class SubSubAlpha extends Alpha
{
    private SubSubAlpha()
    {
        s += "subsub ";
    }
    public static void main(String[] args)
    {
        new SubSubAlpha();
        System.out.println(s);
    }
}

Output:

alpha subsub

Explanation: SubSubAlpha extends Alpha! Since the code doesnt attempt to make a SubAlpha, the private constructor in SubAlpha is okay.

Program 4 :

public class Juggler extends Thread
{
    public static void main(String[] args)
    {
        try
        {
            Thread t = new Thread(new Juggler());
            Thread t2 = new Thread(new Juggler());
        }
        catch (Exception e)
        {
            System.out.print("e ");
        }
    }
    public void run()
    {
        for (int i = 0; i < 2; i++)
        {
            try
            {
                Thread.sleep(500);
            }
            catch (Exception e)
            {
                System.out.print("e2 ");
            }
            System.out.print(Thread.currentThread().getName()+ " ");
        }
    }
}

Output: No Output
Explanation: In main(), the start() method was never called to start ”t” and ”t2”, so run() never ran.
For detail: See Multithreading in Java

Program 5 :

class Grandparent
{
    public void Print()
    {
        System.out.println("Grandparent's Print()");
    }
}

class Parent extends Grandparent
{
    public void Print()
    {
        System.out.println("Parent's Print()");
    }
}

class Child extends Parent
{
    public void Print()
    {
        super.super.Print();
        System.out.println("Child's Print()");
    }
}

public class Main
{
    public static void main(String[] args)
    {
        Child c = new Child();
        c.Print();
    }
}

Output: Compiler Error in super.super.Print()
Explanation: In Java, it is not allowed to do super.super. We can only access Grandparent’s members using Parent. See Inheritance in Java

This article is contributed by Pavan Gopal Rayapati. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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