CODEWITHSUNDEEP
javagenerics
peter.petrov
peter.petrov

Reputation: 39437

Bounded type parameters - is upper bound included?

Reading this here...

https://docs.oracle.com/javase/tutorial/java/generics/bounded.html

"To declare a bounded type parameter, list the type parameter's name, followed by the extends keyword, followed by its upper bound, which in this example is Number. Note that, in this context, extends is used in a general sense to mean either "extends" (as in classes) or "implements" (as in interfaces)."

... but after doing a few tests it seems the upper bound is included. So I think the part in bold is not quite precise. The general sense here is "extends or same type" or "implements or same type".

Is my understanding correct?

To give a student-like example... Say I have this method 

    public static <U extends Animal> void inspect(U u){
        System.out.println("U: " + u.getClass().getName());
    }

Then it seems I can well pass in an Animal object to it i.e. the argument doesn't need to be a Dog, or Cat or any actual subclass of Animal, it can well be just an Animal (I am assuming that Animal is not abstract of course).

Upvotes: 0

Views: 604

Answers (1)

Andreas
Andreas

Reputation: 159076

The full quote is (emphasis mine):

There may be times when you want to restrict the types that can be used as type arguments in a parameterized type. For example, a method that operates on numbers might only want to accept instances of Number or its subclasses. This is what bounded type parameters are for.

To declare a bounded type parameter, list the type parameter's name, followed by the extends keyword, followed by its upper bound, which in this example is Number. Note that, in this context, extends is used in a general sense to mean either "extends" (as in classes) or "implements" (as in interfaces).

To summarize: When the upper bound is Number, the type parameter accept instances of Number or its subclasses.

In short: The upper bound is inclusive.

Upvotes: 2

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