Re: (Class>) casting troubles

[Sebastiaan van Erk]

Zappaterrini, Larry wrote:

Sebastiaan,

Point 1 is a good one. I haven't puzzled that through completely. Upon
initial inspection it seems that it is just the compiler being pedantic
about a scenario that wouldn't arise in practice. I'll have to think
about it some more.

I might be missing something with point 2, but what is wrong with
Class<Foo> clazz = Foo.class?

Foo is a raw type. References to generic type Foo<T> should be parameterized

Regards,
Sebastiaan

If Foo is a generic type (as in the example I gave), then the above assignment will give you a warning (which I don't know how to get rid off without a suppress)...

Cheers,
Larry

-----Original Message-----

From: Sebastiaan van Erk [mailto:[EMAIL PROTECTED] Sent: Saturday, June 07, 2008 3:57 AM

To: users@wicket.apache.org
Subject: Re: (Class<? extends Page<?>>) casting troubles

Zappaterrini, Larry wrote:

Sorry, I should have been more clear about subtype. :) When dealing

with

raw types, the raw type is considered a super type of the generic

type.

So Bar is a super type of Bar<?>. Since RawType extends the raw type
Bar, consider it to be a peer of Bar<?>. When you consider them as
peers, a warning is warranted. The new example you use works due to
erasure. Bar<?> as declared in source code becomes Bar in byte code.

So

the statement becomes:

        Bar bar = new RawBar();

Which is perfectly legal. I have found that most of perceived
inconsistencies in Java generics stems  from erasure and sub type
substitution. The golden rule of generics is that the byte code

produced

by compiling generics will never produce an invalid cast so long as

the

code does not produce any warnings. This causes some things that may
seem intuitive to be illegal.

you though...





Regards,
Sebastiaan

Thanks for your explanation. I still think it's all rather horrible though. Type erasure was a huge mistake if you ask me. Two questions for 1) Can you come up with an example where assigning a Foo<? extends Bar> to a Foo<? extends Bar<?>> causes an invalid cast? (So I can understand why this intuitive seeming assignment is illegal). 2) How do you get rid of the warning in Class<Foo> clazz = Foo.class without using Class<?>? Because it would seem strange if there is no warning free way to use a certain language construct...

-----Original Message-----

From: Sebastiaan van Erk [mailto:[EMAIL PROTECTED] Sent: Friday, June 06, 2008 4:16 PM

To: users@wicket.apache.org
Subject: Re: (Class<? extends Page<?>>) casting troubles

Zappaterrini, Larry wrote:

In the example you have detailed, RawBar is not a subtype of Bar<?>
since it extends the raw type Bar.

I guess it depends on the definition of subtype. It is at least the

case

that the following assignment compiles without warnings (without warnings about unchecked casts):

   Bar<?> bar = new RawBar();

So is it then a subtype? Or isn't it? It's all terribly inconsistent

if

you ask me. :-(

Regards,
Sebastiaan

-----Original Message-----

From: Sebastiaan van Erk [mailto:[EMAIL PROTECTED] Sent: Friday, June 06, 2008 11:31 AM

To: users@wicket.apache.org
Subject: Re: (Class<? extends Page<?>>) casting troubles

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:

   Bar<?> bar = new RawBar(); // works, because RawBar is a subtype

ISSUE 1: Foo<? extends Bar<?>> is not assignable from a Foo<RawBar> where RawBar extends Bar as a raw type. That is, given:

of

Bar<?>

But:

   Foo<? extends Bar<?>> rawbar = new RawBar(); // DOES NOT WORK -

THIS

IS CAUSING ONE HALF OF ALL OUR HEADACHES

(correction:)
    Foo<? extends Bar<?>> rawbar = new Foo<RawBar>(); // DOES NOT

WORK

-

THIS IS CAUSING ONE HALF OF ALL OUR HEADACHES

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

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.

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:


compiles, but generates a warning (reference to raw type). But if

Class<Foo> fc = Foo.class;

you

type this in eclipse:

x fc = Foo.class;

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#1

5.8.2).

So what the heck is the proper type for the class literal??? I

couldn't

find any!

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

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

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).

Regards,
Sebastiaan

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