Question

I have an interface

interface x {
    A getValue();
}

and the implementation

class y implements x {
    public B getValue() { return new B();}
}

B is a subclass of A. This works because of covariant overriding, I guess.

But if I rewrite the interface as

interface x{
    <T extends A> T getValue();
}

I get a warning in the implementation that

Warning needs a unchecked cast to
conform to A.getValue()

What is the difference between 2 versions of the interface? I was thinking they are the same.

Answer 1

The second version of the interface is wrong. It says that the getValue will return any subclass you ask for – the return type will be inferred based on the expression on your left hand.

So, if you obtain a reference to an x (let’s call it obj), you can legally make the following call without compiler warnings:

x obj = ...;
B b = obj.getValue();

Which is probably incorrect in your example, because if you add another class C that also extends A, you can also legally make the call:

C c = obj.getValue();

This is because T is not a type variable belonging to the interface, only to the method itself.