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

Output of Java program | Set 27

Pavan Gopal Rayapati

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Output of Java Program | Set 20 (Inheritance)

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