Re: package protected and compile error

From:
"Daniel Pitts" <googlegroupie@coloraura.com>
Newsgroups:
comp.lang.java.programmer
Date:
3 Nov 2006 16:53:52 -0800
Message-ID:
<1162601632.755523.275990@i42g2000cwa.googlegroups.com>
gk wrote:

code-1
-------------

package packageX;

public class SuperclassX {
    protected void superclassMethodX() {
    }

    protected int superclassVarX;
}

code-2:
------------
package packageY;

import packageX.*;
public class SubclassY extends SuperclassX
  {
        SuperclassX objX = new SubclassY();
        SubclassY objY = new SubclassY();
        void subclassMethodY()
            {
                objY.superclassMethodX();
                int i;
                i = objX.superclassVarX; // compile error is here
           }

  }

Not understanding why there is compile error ?
objX is a type (reference to the Subclass object ) and compiler is
always happy to check with the type .....the objX type really indeed
have superclassVarX as a protected member ....and so there should
not have been any compile error.

I am getting a compile error which is not understandable.

please explain .

thank you


Think of it this way.

You have classX, classY extends classX, and classZ extends classX

classX x = new classX();
classX y = new classY();
classX z = new classZ();

should classY be able to access an object of type classZ's protected
data? I don't think so.
In order for classY to access the protected member of another instance,
that other instance has to be classY. This also has to be visible to
the compiler (either through a cast, or by having the reference
declared as type classY), otherwise the compiler doesn't know if you're
accessing a protected member of a class you are descendant from.

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