Output of Java Program | Set 20 (Inheritance)

Last Updated : 14 Aug, 2019

Prerequisite - Inheritance in Java Predict the output of following Java Programs. Program 1 :

JAVA

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 :

JAVA

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 :

JAVA

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 :

JAVA

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 :

JAVA

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 Related Article: Quiz on Inheritance in Java