"default" access and classpath

[meera kanekal]

I am trying to understand the concept of specifying the fully qualified class name when trying to access a class with default access in one package by its subclass in another package.
So I have the following:

/Users/meera/SCJP/CH1/src/Package1/Accesslevel.java

package Package1;
 
class AccessLevel {
String s;
public String returnS(){
return s = "Hello";
}
 
}

/Users/meera/SCJP/CH1/src/Package2/TestAccess.java

package Package2;
 
//import Package1.AccessLevel;
public class TestAccess extends 
/Users/meera/SCJP/CH1/src/Package1.AccessLevel{
 
/**
 * @param args
 */
public static void main(String[] args) {
// TODO Auto-generated method stub
 
}
 
}

According to K&B SCJP5, Chapter1 page 14 the only way to access a class with default access in a different package is to do one of the following:
1) Make the class public
2) Use fully qualified name.
But my above code is not compiling. So my fully qualified name is not being recognized. So what is wrong with the name because that is the correct directory. Please help.
Thanks,
Meera


[HENRY: Added Code tags]
[ December 02, 2008: Message edited by: Henry Wong ]

[Duc Vo]

Originally posted by meera kanekal:

2) Use fully qualified name.

No, it should be extending the class. In your code, it should be either

package Package2;
 
public class TestAccess extends Package1.AccessLevel{
 
/**
 * @param args
 */
public static void main(String[] args) {
// TODO Auto-generated method stub
 
}
 
}

OR

package Package2;
 
import Package1.AccessLevel;
public class TestAccess extends AccessLevel{
 
/**
 * @param args
 */
public static void main(String[] args) {
// TODO Auto-generated method stub
 
}
 
}

[ December 02, 2008: Message edited by: Duc Vo ]

[M Srilatha]

This code doesnt compile because the class AccessLevel has default access level. That means its not visible outside its package. Even if we use fully qualified name it doesnt compile.

And i guess according to K&B(page 14), to make the above code work, one of the following two things can be done:
1. Having those two classes in the same package
2. Making the class AccessLevel public.

Hope its clear!

[meera kanekal]

Srilatha that was precisely my point. I want to access the class using the fully qualified name.
Duc Vo I tried both the things you suggested earlier and it still does not compile. That is why I posted it here hoping to find out how to use fully qualified name for accessing the class. Thank you both for your replies.
Thanks,
Meera

[Ankit Garg]

If a class has package visibility, then you cannot access it outside it's package in any ways. Also please post the page number in the book where you found this written that you can access it using the fully qualified class name. It might be an error in the book...

[Duc Vo]

Originally posted by meera kanekal:
Srilatha that was precisely my point. I want to access the class using the fully qualified name.
Duc Vo I tried both the things you suggested earlier and it still does not compile. That is why I posted it here hoping to find out how to use fully qualified name for accessing the class. Thank you both for your replies.
Thanks,
Meera

Oops, sorry. Didn't read your post properly. It's right, it wouldn't work. Have to be in the same package.

[meera kanekal]

Ankit I have posted the page number in my post but in case you missed it, it is in K&B SCJP5, Chapter1 page 14. I do not think it is an error. They specifically mention that there are 2 solutions to this problem.
1)Make the superclass public in its package
OR
2) Use the fully qualified name.
I was trying to understand how to use the second solution.
Thanks,
Meera