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.
-----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:
>>
>> ISSUE 1: Foo<? extends Bar<?>> is not assignable from a Foo<RawBar>
>> where RawBar extends Bar as a raw type. That is, given:
>>
>> 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
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:
>>
>> Class<Foo> fc = Foo.class;
>>
>> compiles, but generates a warning (reference to raw type). But if 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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