Raw type. References to generic types should be parameterized

Question

I have a Cage class:

public class Cage {
// the construtor takes in an integer as an explicit parameter
...
}

I am

Answer 1

Cage is a generic type, so you need to specify a type parameter, like so (assuming that there is a class Dog extends Animal):

private Cage cage5 = new Cage(5);

You can use any type that extends Animal (or even Animal itself).

If you omit the type parameter then what you wind up with in this case is essentially Cage. However, you should still explicitly state the type parameter even if this is what you want.