Raw to <? extends Object>

Question

I just don't understand this.

List list = new ArrayList();
List <? extends Object> list1 = list; // unchecked conversion warning.

Since Object is the highest upper bound in Java, I don't see any reason why the warning is there.

Update 1:

With regard to akf's answer:

I understand perfectly what you are saying. I already know that. But <? extends Object> is the highest upper bound. Which means you are have any type too you want. Basically <?> == <? extends Object>.

You can try this on your code and you will see <?> == <? extends Objects>

Update 2:

With regard to Sheng's answer:

List  list = new ArrayList ();
List.add("a");
List <? extends Runnable> list1 = list; //warning here

Why no warning here?

List <?> list2 = list; // No warning here

Update 3:

I'm just revisiting the above and still puzzled.

Since the following is permitted by the compiler:

1. List a = new ArrayList();

2. List <?> b = a;

3. List <? extends Object> c = a; // with warning of course

   for (Object obj : b) {}
   // I don't agree with your statements above that <?> cannot be 
   
   // written in the for (:) loop as shown here
   for (Object obj : c) {}
   

Both are permissible. So i still don't see why the unchecked warning when assiging raw to <? extends Object>

Answer 1

This question, and in particular this answer, have some more details on the differences between ? and ? extends Object. I still haven't found anything that says why you get a warning assigning from List to List<? extends Object>, though.

Answer 2

If I'm thinking correctly, then by default the compiler assumes you mean this

List<?> list = new ArrayList();

The ? means that you can have any generic type you want. This is why

List list = new ArrayList();
List <?> list2 = list

works, because for the compiler they are the same thing

However, when you do this

List<?> list = new ArrayList();
List<? extends Object> list2 = list

You're limiting its scope. Because you are limiting the scope, you get a warning. Yes, I know that you don't think you are, but to the compiler you are. If you're absolutely sure you know what your doing, just ignore it or suppress it