Access Modifier/Package

[Nabila Mohammad]

This is from an online Mock Exam - http://www.akgupta.com/Java/mock_exam.htm

 // Super.java
 
 package p1 ;
 public class Super {
    protected void say ( String s ) {
      System.out.println ( s ) ;
    }
 }
 
// SubOne.java
 
package p1 ;
public class SubOne extends p1.Super { }
 
// SubTwo.java
 
1. package p2 ;
2. public class SubTwo extends p1.Super {
3.     public static void main ( String [ ] args ) {
4.         p1.Super s = new p1.Super() ;
5.         p1.SubOne s1 = new p1.SubOne () ;
6.         SubTwo s2 = new SubTwo() ;
7.         s.say (" Super ") ;
8.         s1.say ("SubOne") ;
9.         s2.say ("SubTwo") ;
10.     }
11. }   

The say() method defined in Super.java has a protected modifier
That would mean it's accessible within the it's own package and in the subclass which may be outside the package.

But i still get the error in line 7 and 8 , something to do with protected modifier.

I think I am missing something ..just dont know what!

[Harshit Rastogi]

Hi
The line 7 and 8 gives error because you can acces a protected method only via class which has inherited it and not by directly creating a object of a class in a different package.

[vidhya suvarna]

The code is giving Compiler Error because the method say() in Super is available to SubTwo & SubOne via inheritance though its access is protected.
So you do not need any reference to access them, they can be access just be their name.

[Harshit Rastogi]

Originally posted by Harshit Rastogi:
Hi
The line 7 and 8 gives error because you can acces a protected method only via class which has inherited it and not by directly creating a object of a class in a different package.

Let me rephrase it
The line 7 and 8 gives error because you can acces a protected method only via class object (SubTwo) which has inherited it and not by directly creating a object of a class directly (Super, SubOne)in a different package.

[Nabila Mohammad]

I have slightly modified the code.

 // Super.java
 
 package p1 ;
 public class Super {
    protected void say (  ) {
      System.out.println ( "Super" ) ;
    }
 }
 
// SubOne.java
 
package p1 ;
public class SubOne extends p1.Super {     
protected void say (  ) {
      System.out.println ( "SubOne ) ;
}
 
// SubTwo.java
 
 package p2 ;
 public class SubTwo extends p1.SubOne {
        protected void say (  ) {
      System.out.println ( "SubTwo ) ;
   public static void main ( String [ ] args ) {
         SubTwo s2 = new SubTwo() ;
         s2.say ("SubTwo") ;
    }
 }   

Thanks.
I kind of rememeber something about accesing protected members through inheritance and not by reference...
But i still can't the right statement for acessing say() in Super and SubOne class.

If i could get some help with that....
[ September 11, 2008: Message edited by: Nabila Mohammad ]

[vidhya suvarna]

For the code you provided at the start, you can access the say() of Super by say("Java")...

[Paul Somnath]

@NABILA MOHAMMAD
Remember that a protected class members can be always accessed thorough inheritance but not through references.
Have a look at the following code:

//SubOne.java
package p1;
 
public class SubOne extends Super
{
protected int x = 5;
protected void say()
{
System.out.println("In SubOne");
}
public void hello()
{
System.out.print("In Public Method in SubOne");
}
}
 
//Super.java
package p1;
 
public class Super
{
protected String str = "Super Class Member";
protected void say()
{
System.out.println("In Super");
}
}
 
//SubTwo.java
package p2;
 
import p1.*;
 
public class SubTwo extends SubOne
{
    public static void main ( String [] args ) 
    {
    p1.Super s = new p1.Super();
    SubOne s1 = new SubOne();
    SubTwo s2 = new SubTwo() ;
    s2.Method();
    s2.say ();
    
    //s1.x = 10; // Error!!, cannot be accessed through reference.
    //s1.say();//Error!!, same reason as above. So methods cant be accessed through reference.
    }
    protected void Method()
    {
    System.out.println(x);//OK, since accessed through inheritance
    System.out.println(str);//Can access Super members in this way.
    }
}

So the say method cannot be accessed through reference.
Can you spot the difference now?
For more explanations, goto K&B: page 34-39..

Have a look at this url for more.
[ September 12, 2008: Message edited by: Paul Somnath ]

[Nabila Mohammad]

Originally posted by Paul Somnath:
@NABILA MOHAMMAD
Remember that a protected class members can be always accessed thorough inheritance but not through references.
Have a look at the following code:

//SubOne.java
package p1;
 
public class SubOne extends Super
{
protected int x = 5;
protected void say()
{
System.out.println("In SubOne");
}
public void hello()
{
System.out.print("In Public Method in SubOne");
}
}
 
//Super.java
package p1;
 
public class Super
{
protected String str = "Super Class Member";
protected void say()
{
System.out.println("In Super");
}
}
 
//SubTwo.java
package p2;
 
import p1.*;
 
public class SubTwo extends SubOne
{
    public static void main ( String [] args ) 
    {
    p1.Super s = new p1.Super();
    SubOne s1 = new SubOne();
    SubTwo s2 = new SubTwo() ;
    s2.Method();
    s2.say ();
    
    //s1.x = 10; // Error!!, cannot be accessed through reference.
    //s1.say();//Error!!, same reason as above. So methods cant be accessed through reference.
    }
    protected void Method()
    {
    System.out.println(x);//OK, since accessed through inheritance
    System.out.println(str);//Can access Super members in this way.
    }
}

So the say method cannot be accessed through reference.
Can you spot the difference now?
For more explanations, goto K&B: page 34-39..

Have a look at this url for more.

[ September 12, 2008: Message edited by: Paul Somnath ]

Hi Paul,
Thanks for the explanation.
I had already gone through the website as well as the K&B.

However they have both mentioned accessing variable in the Parent class-
which has not been re defined in the child class.

If i have defined a say() method in all the three classes - Super,Subone and SubTwo
Or a variable s in all the three - that case how do I call the variable or method in Super or SubOne class because it will always give the value of SubTwo.I mean is it even possible!?
[ September 12, 2008: Message edited by: Nabila Mohammad ]