Please explain this code

[Sajan Joseph]

I have been doing the exercises in Thinking In Java and came across this one. Can anyone please explain this to me :
package p1;
public class A{// A is in package p1
protected void a(){
System.out.println("Hello");
}
}
package p2;
import p1.*;
public class B extends A{ // B is a sub class of A in another
// package p2

public static void main(String args[]){
A a1=new A();
a1.a();
}
}
The compiler gives me an error message : a() has protected access in class A.
Changing the code in class B to:
A a1=new B() ---> also does not help. Compiler gives the same error message.
The code compiled and ran only when I changed the code :
B a1=new B();
My question is :
As per the defenition of protected, another package subclass can invoke a protected method in its superclass. B is a subclass of A in package p2. Why is it not able to access a() when I had written :
A a1=new A();
a1.a();
OR when I had written :
A a1=new B(); // trying late binding (dynamic method invocation)
a1.a();

Can anybody clarify this to me.
Thank you,
SJ

[Michael Mendelson]

Sajan, this code worked for me...

class A{ // A is in package p1
protected void a(){
System.out.println("Hello");
}
}
class B extends A{ // B is a sub class of A in another
public static void main(String args[]){
A a1=new A();
a1.a();
}
}

I removed the package references for simplicity. I suggest deleting all A.class and B.class files and trying again.
Or, for simplicity's sake, try compling them from the same file.

[This message has been edited by Michael Mendelson (edited January 13, 2001).]

[Sajan Joseph]

Originally posted by Michael Mendelson:
[B]Sajan, this code worked for me...

class A{ // A is in package p1
protected void a(){
System.out.println("Hello");
}
}
class B extends A{ // B is a sub class of A in another
public static void main(String args[]){
A a1=new A();
a1.a();
}
}

I removed the package references for simplicity. I suggest deleting all A.class and B.class files and trying again.
Or, for simplicity's sake, try compling them from the same file.

[This message has been edited by Michael Mendelson (edited January 13, 2001).][/B]

Michael,
I guess for protected access modifier to be tested, you need to put both classes in different packages. What I am not able to understand is why I cannot instantiate an object of class A in class B, and call the protected method a() on the object of A.
Can you clarify that to me .
SJ

[Daniel Olson]

After reading your question, I tried writing class B without the import statement.

public class B extends p1.A

If you do this, the compiler will not recogize the A constructor.

A a1=new A();

So I wonder if the reason the compiler complains "a() has protected access in class A" is that A is coming from the import and considered an outside class, while B is a subclass which can access the protected method.
I like to see an explanation from someone who understands this better.

[Peter Tran]

Sajan,
From RHE 2nd Ed regarding the use of the protected access modifier:

A protected feature of a class is available to all classes in the same package, just like a default feature. Moreover, a protected feature of a class is available to all subclasses of the class that owns the protected feature.
Okay, to your example. I too get the same compiler error when the classes are in different package.

A a1 = new A();
// a1.a();  // Compiler error "a() has protected access in p1.A"

You're trying to access a() outside package p1. This is not allowed, because a() is a protected. Granted that the handle a1 is to the class A which technically owns the method. If you try to acess the method outside the package, the compiler will not allow it.

B b1 = new B();
b1.a();

The reason why the above code works is from the second sentence in RHE. Now you're invoking a() via a B reference. Since B is a subclass of A and has all access to A's public and protected method, you can invoke the method via the clas B reference handle.
That's my interpretation. I'm curious what Michael has to say, since I'm quoting his book. His example works because both classes are in the same package.
-Peter

[Sajan Joseph]

Thanks Peter, Daniel Michael,
I am still not sure I have understood this properly. What I have understood is that a protected member can be accessed in a different package subclass only through the sub class reference, not through the superclass reference.
I do not have any actual coding experience. Is this feature used frequently in programming? What advantage does a class developer get in declaring a member as protected? And how does it affect the client who uses it?
SJ

[hari pillai]

Hi Sajan,
I think the fact that you cannot access the protected super classmethod/field from a subclass in other package is purposely implemented to make the method/field ,well,protected from classes in other packages.
Whoever makes a super class method protected thereby allows only subclasses in the same package to use its methods or manipulate itsfields. Users in other package gets the method/field of the subclass to play with .
This is my belief which is always open for correction.

[Swati Samudra]

If I save class A in package P1 and class B in default package that is current directory. I am able to compile and run the program. Even if I give A a1=new A(); or B a1=new B(). I am not having any problem with this.
But I have another problem. If I save class A in package P1 and class B in another package P2. I am able to compile A.java by keeping my current directory P1. But I am not able to compile B.java. Because if I compile from P2 directory I am getting the message "package P1 does not exist" and if I compile from one-up directory I get "can't read B.java".
How can I compile B.java which is in another package, importing class from A.java which is another package.
I don't know how to do this. My problem is I am not able to compile file in different package which is importing classes from another package. From which prompt should I compile the file B.java.

[Swati Samudra]

Sajan,
I have found the solution to my problem. My classpath was not set to one-up, but one above the one-up in Autoexec.bat, so I was having the problem in compiling B.java. While compiling A.java I was making P1 as my current directory. But while compiling B.java if I made P2 as my current directory it would not able to find P1 package. So I had to set classpath to one-up above the package in my Autoexec.bat file and I was able to compile both the files.
Let me inform you, the program is just fine and runs with any i.e either superclass or subclass reference.
[This message has been edited by Swati Samudra (edited January 18, 2001).]