Re: Generic type cast involving wildcard type

From:

Lew <noone@lewscanon.com>

Newsgroups:

comp.lang.java.programmer

Date:

Wed, 29 Jul 2009 20:06:39 -0400

Message-ID:

<h4qo6g$aca$1@news.albasani.net>

Daniel Thoma wrote:

First of all, I found an interesting workaround:

Base<? extends CharSequence> base4 = new Derived<String>();
Derived<? extends CharSequence> derived4 = cast(base4);

<U> Derived<U> cast(Base<U> base) {
return (Derived<U>) base;
}

This example causes no unchecked warning. I now tried to read the JLS in
more detail and came up with the following explanation:

In the second line on the evaluation of base4 capure conversion is applied
(JLS 5.1.10) which results in a type captureX extends CharSequence].
Now typinference yields U = captureX. Therefore case(base4) has has a return
type of Derived<captureX>. Derived<captureX> is a subtype of Derived<?
extends CharSequence> (JLS 4.10.2), therefore the result can be assigned to
derived4.

As I understand the JLS, capture conversion is never carried out for the
type of derived4.
The direct cast should behave any different. Capture-conversion will be
applied to the cast-expression, resulting in a type Derived<captureX>. The
compiler message supports this, since it only mentions a capture variable
for the "found"-part.
So the question to me is, why is Derived<captureX> assingable to Derived<?
extends CharSequence> in case of the method invokation [sic] but not in case of
the direct cast.

Because the non-wildcard type parameter U captures the exact capture of the
'base' parameter and pins it to the method return's type parameter, and then
when you assign the return value to the wildcarded variable that variable in
turn is able to pick up on this same capture, but in the process effectively
"forgetting" which capture it picked up. I think of the method as "stepping
down" the wildcard to a specific capture, then the assignment of the return
value to a variable as "stepping up" the capture to a wildcard. In between,
the method ensures a clean pass-through of the type.

Without the method the compiler doesn't have enough information to guarantee
the compatibility of the step-down and step-up. I've seen similar idioms in
the literature for Java generics, whereby a helper method nails down the type
parameter.

--
Lew