why compile time error???

[cs singh]

//in file A.java
package p1;
public class A
{
      protected int i = 10;
      public int getI() { return i; }
}
//in file B.java
package p2;
import p1.*;
public class B extends p1.A
{
   public void process(A a)
   {
      a.i = a.i*2;
   }
   public static void main(String[] args)
   {
      A a = new B();
      B b = new B();
      b.process(a);
      System.out.println( a.getI() );
   }
}
    

Answer is: Compile time error.
I'm not getting the reason for it

[Edisandro Bessa]

Hi Singh,

In class A, the member "i" is marked as protected, so it's accessible only by classes defined in the same package.

If you want to make the "i" member accessible through class B, you must either mark "i" as public or default.

[david allen]

What Edisandro wrote is not exactly correct.

>> If you want to make the "i" member accessible through class B, you must either mark "i" as public or default.

Default is more restrictive than protected. You will have to declare "i" as public in this example to get it to compile.

[Keith Lynn]

protected members can be accessed from within subclasses of the class in which they are defined.

However, because in the method process in class B, B is not involved in the implementation of an A object, within its methods we cannot refer to the protected members of A.

There is more explanation of this in the Java Language Specification.

http://java.sun.com/docs/books/jls/second_edition/html/names.doc.html#62587

[ashish kediya]

Originally posted by cs singh:

//in file A.java
package p1;
public class A
{
      protected int i = 10;
      public int getI() { return i; }
}
//in file B.java
package p2;
import p1.*;
public class B extends p1.A
{
   public void process(A a)
   {
      a.i = a.i*2;
   }
   public static void main(String[] args)
   {
      A a = new B();
      B b = new B();
      b.process(a);
      System.out.println( a.getI() );
   }
}
    

Answer is: Compile time error.
I'm not getting the reason for it call it as not through a but through b
because it is protected n it inherit as privetly so it can't call by the class where it reside
b know that it's it's own property

[ashish kediya]

Originally posted by cs singh:

//in file A.java
package p1;
public class A
{
      protected int i = 10;
      public int getI() { return i; }
}
//in file B.java
package p2;
import p1.*;
public class B extends p1.A
{
   public void process(A a)
   {
      a.i = a.i*2;
   }
   public static void main(String[] args)
   {
      A a = new B();
      B b = new B();
      b.process(a);
      System.out.println( a.getI() );
   }
}
    

Answer is: Compile time error.
I'm not getting the reason for it

[ashish kediya]

Originally posted by cs singh:

call it as not through a but through b 
because it is protected n it inherit as privetly so it can't call by the class where it reside
b know that it's it's own property

Answer is: Compile time error.
I'm not getting the reason for it

[bnkiran kumar]

I have one doubt, even in other packages, the fields declared as protected will to available to classes which extends that class which has the protected field, how can this will create complie time error???

[Vidhya Hari]

The accessibilty rule of protected is package+kids(subclass)...Which means subclass members can access its protected superclass members only thru inheritance and not via a reference to the superclass...In the above case,since u've accessed the protected i as a.i(via its reference),it gives a compiler error..Try accessing it directly as i(via inheritance)...It should work...

[S Thiyanesh]

The protected members of the superclass are accessible to the subclass only through inheritance i.e instance of subclass.
Not through the super class instances.

[Sanju Thomas]

I totally agree with Thiyanesh. and the protected members are available in any level of inheritance. The visibilty will not come down to private in any level of inheritance.

[Nikhil Bansal]

Yes I agree with Thiyanesh.

[Sandeep Vaid]

whats Edisandro has written is not correct.

We can access protected members in diffrent packages but only in a subclass (or subclasses or subclass) only through the subclass reference ( and not via superclass reference)

[Arno Reper]

I tryed this :
//in file A.java
//package p1;
public class A
{
protected int i = 10;
public int getI() { return i; }
}

//in file B.java
//package p2;
//import p1.*;
class B extends A //p1.A
{
public void process(A a)
{
a.i = a.i*2;
}
public static void main(String[] args)
{
A a = new B();
B b = new B();
b.process(a);
System.out.println( a.getI() );
}
}

and it compiles...the problem is with the package...
I think the A that B extends isn't the A from p1.* but another A from p1.p1.*

[ March 21, 2006: Message edited by: Reper Arno ]
[ March 21, 2006: Message edited by: Reper Arno ]

[Edisandro Bessa]

Hi Guys,

I don't know what I had in my mind when I wrote this.

In class A, the member "i" is marked as protected, so it's accessible only by classes defined in the same package.

If you want to make the "i" member accessible through class B, you must either mark "i" as public or default.

For sure it's not correct.

[Edisandro Bessa]

Hi,

I'm not sure but I think you cannot explicitly access a protected member through an instance variable when you're outside the protected member class's package.

But about one thing I am sure, you can access through inheritance.

I tried to change the process method to ...

public void process(A a){      
    super.i = super.i*2;
 
         [B]//OR[/B]
    
    i = i * 2;
    }

Once I ignored the (a) argument, I'm not using it in anywhere else, of course that the result is not the same as expected from the original code but this new one proves that you cannot access a protected member through an instance variable when you're outside the protected member class's package.

I think it makes sense.

Any comments are welcome.