Re: (Class>) casting troubles

[Sebastiaan van Erk]

ARgh, you always make typos with this stuff.

See correction.

Sebastiaan van Erk wrote:

Martin Funk wrote:

Class<? extends Page<?>> means "class of (anything that extends (page of
anything))".

I'm not so sure.

There are 2 separate issues:



  static class Foo<T> {
  }

  static class Bar<T> {
  }

  static class RawBar extends Bar {
  }

  static class SubBar<T> extends Bar<T> {
  }

Thus:



But:

ISSUE 1: Foo<? extends Bar<?>> is not assignable from a Foo<RawBar> where RawBar extends Bar as a raw type. That is, given: Bar<?> bar = new RawBar(); // works, because RawBar is a subtype of Bar<?> Foo<? extends Bar<?>> rawbar = new RawBar(); // DOES NOT WORK - THIS IS CAUSING ONE HALF OF ALL OUR HEADACHES

(correction:)


Btw, this does work (like you expect):
   Foo<? extends Bar<?>> rawbar2 = new Foo<SubBar<?>>();

Foo<? extends Bar<?>> rawbar = new Foo<RawBar>(); // DOES NOT WORK - THIS IS CAUSING ONE HALF OF ALL OUR HEADACHES

Note that this is the issue that complete baffles me, as RawBar is a subtype of Bar<?>, so I *really* *really* *REALLY* have no idea why the compiler chokes on this.

x fc = Foo.class;

ISSUE 2: The class literal of a generic type returns a class of a raw type. Thus Foo.class return Class<Foo>. This is also really messed up, because: Class<Foo> fc = Foo.class; compiles, but generates a warning (reference to raw type). But if you type this in eclipse: and use eclipse quickfix to change "x" to the "correct" type, it'll change it to precisely Class<Foo> (the JLS is very short about this, see also http://java.sun.com/docs/books/jls/third_edition/html/expressions.html#15.8.2).

Finally, note that when you define a method like this:

  static void method1(Foo<? extends Bar<?>> y) {
  }





Regards,
Sebastiaan

So what the heck is the proper type for the class literal??? I couldn't find any! it works like a charm for a new Foo<SubBar<String>>, i.e., the "Foo of (anything that extends (bar of anything))" really is the correct interpretation. It's just that the interaction with raw types is completely *foobar* (pun intended).

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