Q of inner class in different package

[Claire Yang]

I wrote the following codes to test the access modifiers of inner classes:

//File: p1/A.java
package p1;
public class A {
public class Inner1 {
public void m1() { System.out.println("public inner class"); }
}
protected class Inner2 {
public void m1() { System.out.println("protected inner class"); }
}
}
//File: p2/A.java
package p2;
import p1.*;
public class A extends p1.A {
public static void main(String args[]) {
p1.A a1=new p1.A();
p1.A.Inner1 i1=a.new Inner1(); //line1
p1.A.Inner2 i2=a.new Inner2(); //line2
} 
}

When compiled, line1 is ok, but line2 causes a compile error: Inner2() has protected access in p1.A.Inner2(both the code--remove package&import lines--are ok if they were put in the same directory). Where am I wrong? Can we use inner class as general class members? Or using inner class like this is meaningless?

[Alfred Kemety]

Well member inner classes are like other class member variables, methods, thus to access a protected member from a subclass of another package, you have to access it through an instance of the subclass not the super class.
Thus if you change your code to the following, it should work just fine.

//File: p2/A.java
package p2;
import p1.*;
public class A extends p1.A {
public static void main(String args[]) {
p1.A a1=new p1.A();
p1.A.Inner1 i1=a1.new Inner1(); //line1
A a2=new A();
p1.A.Inner2 i2=a2.new Inner2(); //line2
} 
}

HTH

[Claire Yang]

Thanks for your reply, Alfred.
I modified the code as you said, I still got the same compile error:

p2\A.java:9: Inner2() has protected access in p1.A.Inner2
                p1.A.Inner2 i2=a2.new Inner2(); //line2
                                  ^

I compiled the 2 files in their parent directory with the following commands:
javac p1\A.java
javac p2\A.java
I think it's right.