Question

I’m trying to figure out why I can’t I use super in a generic method declaration? See the example below, it doesn’t compile, but I’m not sure why.

public class A<E> {
    public <T extends A> void test1(T t) {   // Compiles fine
    }
    public <T super A> void test2(T t) {     // Compilation error??
    }
    public <T super A> <T> void test3(T t) { // Compilation error??
    }
}

class A {}
....

Answer 1

Think about it, if a method could have the signature

public <T super A> void test2(T t) {
}

then it could take any object as argument. Why? Because, for example, you can always infer T as Object (Object is a supertype of A), and any object is an instance of Object, so it works. So declaring it this way would be pretty redundant. You could just write

public void test2(Object t) {
}

for the same effect.