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From: Steven Simpson <ss@my.domain.invalid>
Newsgroups: comp.lang.java.programmer
Subject: Re: Java 8 Lambda with generic types
Date: Wed, 31 Aug 2016 17:19:08 +0100
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On 31/08/16 06:25, Lew wrote:
> On Sunday, August 28, 2016 at 7:15:11 AM UTC-7, Steven Simpson wrote:
>> On 28/08/16 12:46, Arne Vajhøj wrote:
>>> public class GenFun {
>>>      public static interface Toy<E> {
>>>          public List<E> get();
>>>      }
>>>      public static <E> void checkToy(Toy<E> toy) {
>> I found that worked too, but what is the difference between that and
>> <?>, when E is only referenced once?
> When you use a wildcard in lieu of the generic type parameter, you are losing the information that the type in the <?> is the same as `E`. There is no way for the compiler to know that the type of the `Toy<?> toy` argument is compatible with `E`.

It's possible I've conflated Arne's portion of the discussion with my 
own (my mistake), so I'm not sure which E you're referring to.

I mean, if we have:

   interface Toy<F> { List<F> get(); }

   static <E> void checkToy1(Toy<E> toy) { }
   static void checkToy2(Toy<?> toy) { }

   static List<Integer> makeList() { ... }

...why does the compiler accept checkToy1(() -> makeList()), but not 
checkToy2(() -> makeList())?

Is there a simpler case to demonstrate the distinction?  Is there one 
without lambdas/method references?

-- 
ss at comp dot lancs dot ac dot uk