Modifier protected does not hide inherited members to (static) code of superclass package

[Alexander Exner]

K&Bs Book tells me on Page 36:
Once a Subclass-outside-the-package inherits the protected member, that member (as inherited by the subclass) becomes private to any code outside the subclass, with the exception of subclass of the subclass.

I did ask myself what happens if a class in package B extends a class from package A and another (e.g. static) method within package A tries to use some protected stuff from B:

package A;

public class AClass {
  protected void m() {
    System.out.println("A");
  }
}

package B;

import A.AClass;

public class BClass extends AClass {
  protected void m() {
    System.out.println("B");
    System.out.println("Super:");
    super.m();
  }

package A;

import B.BClass;

public class Test {
  public static void main(String[] args) {
    BClass b = new BClass();
    ((AClass)b).m(); // Does compile and return B\nSuper:\nA
    b.m(); // Does not compile
  }
}

I would have assumed that doing ((AClass)b).m(); is ok, because BClass extends AClass and AClass has method m(). m() from BClass should be invisible for "any code outside the subclass", therefore output should be perhaps "A". In fact output is "B\nSuper:\nA".

Please can anybody explain java to me?

KR Alex

[Garg Sharad]

The Output of below statement is correct because
((AClass)b).m(); // Does compile and return B\nSuper:\nA

There are two things happening

one is compile time polymorphism.
It's happening because you are doing BClass object type cast with AClass object so this thing checked at compile time.

second run time polymorphism.
At run time even you are using sub class object with super class reference it'll call sub class object method.



You are getting compile time error at

b.m(); // Does not compile

line because method m access modifier is protected and you can not use out of protected member out side the package without using inheritence.

[Alexander Exner]

Thanks for your clarification, looking at compile time and run time separate explains behavior.

[Mohit G Gupta]

i have posted my query in new post as previous post was resolved
i have changed the code a bit.

AClass.java

 package A;  
  
public class AClass {  
  protected void m() {  
    System.out.println("A");  
  }  
}     

BClass.java

  package B;  
  
import A.AClass;  
  
public class BClass extends AClass {  
private int m=12;
  public void m1() {  
    System.out.println("B");  
    System.out.println("Super:");  
    super.m();  
  }  }  

Test.java

  package A;  
  
import B.BClass;  
  
public class Test {  
  public static void main(String[] args) {  
    BClass b = new BClass();  
    ((AClass)b).m(); //
    b.m1(); //
    //System.out.println("bmvalue"+b.m);
    b.m();
  }  
}    

OUTPUT:
A
B
Super:
A
A

K&Bs Book tells me on Page 36:
Once a Subclass-outside-the-package inherits the protected member, that member (as inherited by the subclass) becomes private to any code outside the subclass, with the exception of subclass of the subclass.

here,m is protected member of AClass that is inherited by BClass.so,by above statement m in BClass would be Invisible outside BClass.
but ,it is accessible in Test.java

SEE LINE 8,11

how is this possible ?

[Abimaran Kugathasan]

Garg Sharad wrote:

one is compile time polymorphism.
It's happening because you are doing BClass object type cast with AClass object so this thing checked at compile time.

Compile time polymorphism is for overloaded methods.

Check this JLS regarding the protected modifier.