Wednesday, 31 July 2013

Access Specifier in Method Overriding

As per the rule in overriding , you cannot apply weaker access specifier over a stronger access specifier.Suppose , your parent class has a method Display() with stronger access specifier like Public, then you cannot use weaker access specifier like private or public when you are overriding the Display() method.

Note :-

 Public is the 1st stronger access specifier
 Protected is the 2nd stronger access specifier
 Default is the 3rd stronger access specifier
 Private is the most weaker access specifier

 Always keep in mind about the access specifiers has vital role in inheritance (OOPS concept).

 Example :-

 * @Author Manoj

class AccessTest{
    protected void display(){ //Stronger Access specifier
        System.out.println("Hello AccessTest:Display");

class TestWithMain extends AccessTest{
    // I am trying to override with weaker specifier , it show error
    // You can use public or protected (higher or same level of access specifier)
    private void display(){
        System.out.println("Hello TestWithMain:display");
    public static void main(String str[]){       
        AccessTest acObj=new TestWithMain();

 Error :- display() in TestWithMain cannot override display() in AccessTest;
 attempting to assign weaker access privileges; was protected
    private void display(){


So, now if you change the access specifier to protected or public then it will work properly.

Changed executable code :-

    public void display(){
        System.out.println("Hello TestWithMain:display");

    protected void display(){
        System.out.println("Hello TestWithMain:display");


Hope it will help you.