accessing a protected access instance variable defined at a superclass

[Ruben Guillen]

Dear All

I am preparing myself for the Certification exam, I am going to have it in 1 week and half. So while studying using the SCPJ 6 Study guide book, I stumbled on the following test sample that confuse me.

In chapter 10 the following test is

  
1. package pkgA;
2. public class Foo{
3.    int a = 5;
4.    protected int b = 6;
5. }  

1. package pkgB;
2. import pkgA.*;
3. public class Fiz extends Foo {
4.    public static void main(String[] args){
5.       Foo f = new Foo();
6.       System.out.print("  " + f.a);
7.       System.out.print(" " + f.b);
8.       System.out.print(" " + new Fiz().a);
9.      System.out.print(" " + new Fiz().b);
10.   }
11. }

What is the result? (Choose all that apply.)
A. 5 6 5 6
B. 5 6 followed by an exception
C. Compilation fails with an error on line 6
D. Compilation fails with an error on line 7
E. Compilation fails with an error on line 8
F. Compilation fails with an error on line 9

The correct Answers are: C, D and E are correct. Variable a (default access) cannot be accessed from outside the package. Since variable b is protected, it can be accessed only through inheritance.

The wrong answers are: A, B, and F are incorrect based on the above. (Objectives 1.1, 7.1)

Now the doubt: I am really feeling frustated because this is strange behaviour, my understanding was based on the explanation provided on chapter 1 page 32 to 36.
Why F is not included (I run the test on the computer and it does not consider as an error condition)

Please could you explain me why "new Fiz().b" does not thrown a compilation error as "new Foo().b" does. For me I can not find the logic behind.

Thank for any help provided.

[Lucas Smith]

Protected instance variables can be accessed, BUT only through the inheritance (if they are accessed from within a class which is in other package).
Let's follow that logic:
1). You instantiate the subclass (protected member can be accessed through the inheritance).
2). You can access protected member:
a). anonymously - new SubClass().member;
b). with reference type:
SubClass sc = new SubClass();
sc.member;

YOU ACCESS IT THROUGH THE INHERITANCE - YOU DID NOT DEFINE "member" INSIDE "SubClass" - IT IS INHERITED.

But invoking (from within SubClass):
SuperClass superClass = new SuperClass();
superClass.member; //error
Gives You a compiler error. YOU DO NOT ACCESS "member" THROUGH INHERITANCE.

[Ankit Garg]

Hi Ruben, welcome to javaranch.

This type of question has been asked before like here or here. See if that could help...

[Ruben Guillen]

Dear All

Thank you very much for the prompt response. Lukas explanation was clear for me and I complemented with references provided by Ankit.

It is a relief for me to know this information, one less tricky question to solve during the exam.

..And thanks Ankit for the welcome.

Regards.

[Lucas Smith]

OK, and please have look at the protected static variables. They can be a little trickier.