Question

I was reviewing someone else’s code the other day and I came across a line that raised some concern. To simplify, say I have a generic Class A and an abstract Class B. Is the following instantiation allowed and if so, why?

Object obj = new A<? extends B>();

I personally have never seen an instantiation like the above, although a declaration such as

A<? extends B> obj = null;

would certainly hold. I’ve always used the wildcard in generics to declare method parameters, so I may just not have the experience.

Answer 1

Actually new A<? extends B>() does not compile. It has been consistently illegal since Java 5.

But I guess your original example was something like new A<X<? extends B>>(). The latter is legal in recent versions of Java.

The idea is, when instantiating an object, the value for type parameters can be any non-wildcard type. ? extends B is a wildcard type, so it is disallowed. But X<? extends B> is not a wildcard type, though it has a wildcard type as a component. So you can say legally call new A<X<? extends B>>().

The rules makes sense if you think about it this way. Ultimately it is a byproduct of the more fundamental rule that a wildcard type like ? extends B cannot be the declared type of a field or variable. If A is defined as

class A<T> {
    T value;
}

then the hypothetical new A<? extends B>().value would be a field declared of type ? extends B. Since that is illegal, so is the instantiation. But new A<X<? extends B>>() does not have that problem.