Type mismatch: cannot convert MyClass<E> to MyClass<E>

Question

There is something I am missing about generics and/or inner classes. I want to write a specialized tree class, with a specialized iterator. How can class FooTree have an iterator that returns successive nodes of type FooTree rather than node values of type V? The class starts:

public class FooTree<V>
   private final V value;
   private FooTree<V> left;
   private FooTree<V> right;

I made the Iterator class an inner class since nobody else will care about it. The factory method defined in FooTree to get an iterator is:

public Iterator<FooTree<V>> preorderIterator() {
    return this.new PreorderIterator<FooTree<V>>;
}

Since next() must return a V, as declared in the interface, I added nextNode() to get around the return type. The inner class starts:

private class PreorderIterator<V> implements Iterator<V> {
   private FooTree<V> current;  
   . . .
   public FooTree<V> nextNode() {
      current = FooTree.this;                        **

I get this wonderful compiler error:
Type mismatch: cannot convert from ...FooTree<V> to ...FooTree<V> (both are the same package)
Eh wot!?
I can make it compile by adding a cast: current = (FooTree)FooTree.this; But why should I have to in the first place? Is it trying to be "smart" and figuring the first V might be different from the second V?

What is the right way to do this?

Answer 1

Try making your declaration:

private class PreorderIterator implements Iterator<V>

The <V> in your inner class is unnecessary, and in this case, unhelpful.

What's actually happening is that the <V> in PreorderIterator<V> is "shadowing" the <V> from your outer class, so that although they look like they're the same type parameter, they're actually different parameters that happen to have the same name.