proctected -- Pl HELP

[padmini Babu]

Can anyone help me understand how the answer is correct?
Question ID :957672686580
Consider following two classes:
//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() );
}
}
What will be the output of compiling and running class B ?
Ans: It will not compile
I could not understand the explan too. Pl help
Although, class B extends class A and 'i' is a protected member of A, B still cannot access i , (now
this is imp) THROUGH A's reference because B is not involved in the implementation of A.
Had the process() method been defined as process(B b); b.i would have been accessible as B is involved
in the implementation of B.
Thanks in advance
padmini

[Tamanna Mittal]

Hi,
It's working fine on my machine and i am getting the output as 20. which jdk version are u using? Coz i am using jdk1.3.
------------------
Cheers
Tamanna :-)

[vineetsab]

you have to compile both the files to make it work. If only B.java is compiled then it won't be able to access method in package A, hence the compile error. :>

[Patricia Anderson]

I too have the same doubts about the JQplus answer. I was able to compile it too.

[Anshul Manisha]

Hi Padmini,
I too had a similar kind of situation earlier and the good old bartender Jane Gristi helped me with the concept. Let me first quote the bible of Java here.

6.6.2 Details on protected Access
A protected member or constructor of an object may be accessed from outside the package in which it is declared only by code that is responsible for the implementation of that object.

and then they provide an example of the same

6.6.7 Example: protected Fields, Methods, and Constructors
Consider this example, where the points package declares:

package points;
public class Point {
protected int x, y;
void warp(threePoint.Point3d a) {
if (a.z > 0)// compile-time error: cannot access a.z
a.delta(this);
}
}

and the threePoint package declares:

package threePoint;
import points.Point;
public class Point3d extends Point {
protected int z;
public void delta(Point p) {
p.x += this.x;// compile-time error: cannot access p.x
p.y += this.y;// compile-time error: cannot access p.y
}
public void delta3d(Point3d q) {
q.x += this.x;
q.y += this.y;
q.z += this.z;
}
}

which defines a class Point3d. A compile-time error occurs in the method delta here: it cannot access the protected members x and y of its parameter p, because while Point3d (the class in which the references to fields x and y occur) is a subclass of Point (the class in which x and y are declared), it is not involved in the implementation of a Point (the type of the parameter p). The method delta3d can access the protected members of its parameter q, because the class Point3d is a subclass of Point and is involved in the implementation of a Point3d.
The method delta could try to cast (�5.5, �15.16) its parameter to be a Point3d, but this cast would fail, causing an exception, if the class of p at run time were not Point3d.
A compile-time error also occurs in the method warp: it cannot access the protected member z of its parameter a, because while the class Point (the class in which the reference to field z occurs) is involved in the implementation of a Point3d (the type of the parameter a), it is not a subclass of Point3d (the class in which z is declared).

For all you guys whose code compiles, I suppose that your classes are not in different packages. Please check your code again and let me know.
thanks

[shankar vembu]

hi,
its not compiling for me too.
shanks

[krishna avrs]

Hi all,
Manisha is correct. Code will not compile. In simple words the rule is
A subclass in another package can only access protected members in superclass via references of it's own type(new A().i) or a subtype (new B().i).
It is not possible to access it by reference to superclass.
In this case of A a = new B() - a becomes reference to the superclass for B.
Cheers!
SriKi.