Skip to main content

Creating Copy of the Existing Object in Java Using Clone Method

  In Java, The "Object" class is the root class for all java classes, It means by default all the java classes inherit the Object class. The Object class have some useful methods for performing operation on the class object like clone(), getClass(), equals(), toString(), wait(), etc..,
  The clone() methods is used for creating copy of the existing object. It's means it allocated the new memory for object with values and methods present in which object you are going to copy. For performing this clone() operation you have to implement the Cloneable interface and you have to override the clone() method. See the below example.

public class CloneDemo implements Cloneable {
public int testValue;
public CloneDemo(int value) {
this.testValue = value;
}
//Override the clone() method 
protected Object clone() {
Object newObject = null;
try{
newObject = super.clone();
}
catch(CloneNotSupportedException e){
e.printStackTrace();
}
return newObject;
}
//Main mthod
public static void main(String[] args) {
CloneDemo obj1 = new CloneDemo(10);
CloneDemo obj2 = (CloneDemo) obj1.clone();
System.out.println("The integer value present in object1 : " + obj1.testValue);
System.out.println("The integer value present in object2 : " + obj2.testValue);
}
}

  Here I have create the CloneDemo class and implemented Cloneable interface and override the clone() method for getting copy of the class. In my main method I have created the new object for my class "CloneDemo" and I created the copy of the object1 by calling obj1.clone() method. Finally I printed the value of integer "testValue" present in both objects. The final result should be like this only.

The integer value present in object1 : 10
The integer value present in object2 : 10


Note:
  Hey don't assume the obj1 and obj2 are equals because the obj2 is the copy the obj1, so both objects are not equals. If you compare these two object using equals() method that will returns false.

Comments

Popular posts from this blog

Ext JS 4 – Creating web page using Ext.container.Viewport class and region propery

The Ext.container.Viewport is used for creating general web page and region property is used for splitting web page into different parts. Ext.container.Viewport is a specialized container represents the viewable area of the application (the browser area). It render itself to the document body, there is no need for providing renderTo property and it automatically sizes itself to size of the browser viewport. The viewport also re-size the child elements based on view area(based on browser width and height). The default layout of the viewport is border layout and we can customize this property according to our requirements. The viewport does not provide any scrolling. If necessary, the child elements(generally panels) within viewport needs to provide a scroll feature by using autoScroll property. See the below example for better understanding. This is Home.js file placed into app/view folder. Ext.define('MyApp.view.Home', { extend : 'Ext.container.Viewport&#

Getting servlet init parameter that is defined in web.xml

1. Create a new dynamic web project using eclipse. 2. Create a one new servlet and include the following code. package com.controller ; import java.io.IOException; import javax.servlet .ServletConfig; import javax.servlet .ServletException; import javax.servlet .annotation.WebServlet; import javax.servlet .http.HttpServlet; import javax.servlet .http.HttpServletRequest; import javax.servlet .http.HttpServletResponse; @ WebServlet ( "/LoginController" ) public class LoginController extends HttpServlet {                 private static final long serialVersionUID = 1L;                 public LoginController() {         super ();     }     protected void doPost( HttpServletRequest request, HttpServletResponse response) throws ServletException , IOException {                                 ServletConfig config = getServletConfig ();                                 System.out.println( "Init parameter user nam

Simple Login Application Using Spring MVC and Hibernate – Part 1

 I hope developers working in web application development might hear about MVC architecture. Almost all technologies provide support for MVC based web application development, but the success is based on many factors like reusable of the code, maintenance of the code, future adaption of the code, etc..,  The success of the Spring MVC is “ Open for extension and closed for modification ” principle. Using Spring MVC the developers can easily develop MVC based web application. We don’t need any steep learning curve at the same time we need to know the basics of spring framework and MVC architecture. The Spring MVC consists of following important components. 1. Dispatcher servlet 2. Controller 3. View Resolver 4. Model Spring MVC - Overview  The overall architecture of Spring MVC is shown here.  1. When “Dispatcher Servlet” gets any request from client, it finds the corresponding mapped controller for the request and just dispatches the request to the corresponding contro