Wildcards and Subtyping on Java

Question

I just started studying Wildcards and Subtyping on Java and try to test what I learn.

Suppose:

Class A { public int y=1; }
Class B extends A { public int x=2; }

In main:

List<B> lb = new ArrayList<>();
lb.add(new B());
System.out.println(lb.get(0).y); //Displays member y of Class A
List<? extends A> la = lb;
System.out.println(la.get(0).y); //Can access member y of Class A

List<A> la1 = new ArrayList<>();
la1.add(new A());
System.out.println(la1.get(0).y); //Displays member y of Class A
List<? super B> lb1 = la1;
System.out.println(lb1.get(0).y); //Cannot access member y of Class A? Why?

I don't understand why I cannot access the member y using Lower Bounded Wildcards while it is possible using Upper Bounds Wildcards. Am I missing something?

Answer 1

Consider

Interface X { ... }
Class A { public int y=1; }
Class B extends A implements X { public int x=2; }

List<? super B> lb1 = la1;

Now, the objects in lb1 could be completely unrelated to A and B as long as they implemented X . There's no way for the compiler to know that lb1.get(0) has a member y .

A more "concrete" example:

Class X { ... }
Class A extends X { public int y=1; }
Class B extends A { public int x=2; }

With this hierarchy objects in lb1 could be of type X which has no member y .

Answer 2

Upper bound of U : Has to be at least a U . Anything U has, you can use.

Lower bound of L : Can be at most L (while staying in its hierarchy), but there's no guarantee that it will be.

Answer 3

In simpler words, you cannot get 'TYPE' from a list which is constructed using keyword 'SUPER'..

System.out.println(lb1.get(0).y); // This wrong as you are expecting type A / B

I guess, the following should work

System.out.println(((A)(lb1.get(0))).y);

As you can only get an OBJECT out of this list... you need to cast it to get your 'TYPE'

Answer 4

只需更新行

((A)lb1.get(0)).y