Question

abstract class Type<K extends Number> {
    abstract <K> void use1(Type<K> k);           // Compiler error (Type parameter K is not within its bounds)
    abstract <K> void use2(Type<? extends K> k); // fine
    abstract <K> void use3(Type<? super K> k);   // fine
}

The method generic type K shadows the class generic type K, so <K> doesn’t match <K extends Number> in use1().The compiler doesn’t know anything usefull about new generic type <K> in use2() and use3() but it is still legal to compile . Why <? extends K> (or <? super K>) match <K extends Number>?

Answer 1

First of all, let’s rewrite it to avoid shadowing:

abstract class Type<N extends Number> {
    abstract <K> void use1(Type<K> k); 
    abstract <K> void use2(Type<? extends K> k); 
    abstract <K> void use3(Type<? super K> k);   
}

In the first method K acts as a type parameter of Type<N extends Number>, thus its value sould comply to the bound of Type‘s N. However, method declaration doesn’t have any restrictions on value of K, therefore it’s not legal. It would be legal if you add a necessary restriction on K:

abstract <K extends Number> void use1(Type<K> k);

In the following methods, the actual type parameter of Type is unknown (?), and K imposes additional bound on it, so that there is nothing illegal in these declarations.

Here is a more practical example with the similar declarations:

class MyList<N extends Number> extends ArrayList<N> {}

<K> void add1(MyList<K> a, K b) {
     a.add(b); // Given the method declaration, this line is legal, but it 
          // violates type safety, since object of an arbitrary type K can be
          // added to a list that expects Numbers
          // Thus, declaration of this method is illegal
}

<K> void add2(MyList<? extends K> a, K b) {
     // a.add(b) would be illegal inside this method, so that there is no way
     // to violate type safety here, therefore declaration of this method is legal
}

<K> void add3(MyLisy<? super K> a, K b) {
     a.add(b); // This line is legal, but it cannot violate type safey, since
          // you cannot pass a list that doesn't expect K into this method
}