Java inner class visibility puzzle

Question

Consider the following case:

public class A {
  public A() { b = new B(); }
  B b;
  private class B { }
}

From a warning in Eclipse I quote that: the java complier emulates the constructor A.B() by a synthetic accessor method. I suppose the compiler now goes ahead and creates an extra "under water" constructor for B.

I feel this is rather strange: why would class B not be visible as a.k.o. field in A? And: does it mean that class B is no longer private at run time? And: why behaves the protected keyword for class B different?

public class A {
  public A() { b = new B(); }
  B b;
  protected class B { }
}

Answer 1

http://www.j2ee.me/docs/books/tutorial/java/javaOO/nested.html and as for a puzzle :-) :

http://blogs.sun.com/darcy/entry/nested_inner_member_and_top

Answer 2

You need to use

this.new B();

Answer 3

Inner classes are essentially a hack introduced in Java 1.1. The JVM doesn't actually have any concept of an inner class, and so the compiler has to bodge it. The compiler generates class B "outside" of class A, but in the same package, and then adds synthetic accessors/constructors to it to allow A to get access to it.

When you give B a protected constructor, A can access that constructor since it's in the same package, without needing a synthetic constructor to be added.

Answer 4

The access of class B and its constructor do not have to be the same. You can have a private inner class with a package-scope constructor, and this is what I usually do.

public class A {
  public A() { b = new B(); }
  B b;
  private class B {
    B() { }
  }
}