Question

This program does not compile:

public class xx {
    static class Class1<C> {
        void method1(C p) {
        }
    }
    static class Class2<T> extends Class1<Class<? extends T>> {
        T object;
        void method2() {
            this.method1(this.object.getClass());
        }
    }
}

The error is:

xx.java:10: method1(java.lang.Class<? extends T>) in xx.Class1<java.lang.Class<? extends T>>
cannot be applied to (java.lang.Class<capture#215 of ? extends java.lang.Object>)
        this.method1(this.object.getClass());

Why does this happen? Why does the compiler seemingly believe that object.getClass() returns Class<? extends Object> instead of Class<? extends T> ?

Answer 1

Object.getClass() is defined to return a Class<? extends |T|>, where T is the statically known type of the receiver (the object getClass() is called on). Take special note of the vertical bars, the erasure operator. The erasure of a type variable is the erasure of its leftmost bound. In your case that’s the implicit bound Object. So you get back a Class<? extends Object>, not a Class<? extends T>.

Why is that?

Imagine T = List<Integer>, you could suddenly do the following without unchecked warning:

List<String> myStrings = new ArrayList<>();
List<Integer> myInts = new ArrayList<>();
List<Integer> myIntyStrings = myInts.getClass().cast(myStrings);
myIntyStrings.add(-1);
String myString = myStrings.get(0); // BANG!

But thankfully we do get a warning.. 😉