Regarding generics

[Venu Chakravorty]

Can't we use the 'super' keyword when defining a generic class or method? Like:

 
class Animal {}

class Dog extends Animal {}

class MyClass {
    public <X extends Animal> void m1(X o) {   //'extends' is fine
	System.out.println(o.getClass());
    }
    public <Y super Dog> void m2(Y o) {   //this causes the error
	System.out.println(o.getClass());
    }
}

class A {
    public static void main(String[] args) {
	MyClass mc = new MyClass();
	mc.m1(new Dog());
    }
}

The compiler replies:
A.java:9: > expected
public <Y super Dog> void m2(Y o) {
^
A.java:9: illegal start of type
public <Y super Dog> void m2(Y o) {
^
A.java:9: '(' expected
public <Y super Dog> void m2(Y o) {
^
3 errors

shell returned 1

Regards, Venu.

[Ulrika Tingle]

venu chakravorty wrote:Can't we use the 'super' keyword when defining a generic class or method?

This is from Angelika Langer's generics site:

"Type parameters can have several bounds, like in class Box<T extends Number> {...} . But a type parameter can have no lower bound, that is, a construct such as class Box<T super Number> {...} is not permitted. Why not? The answer is: it is pointless because it would not buy you anything, were it allowed. Let us see why it is pointless by exploring what a upper bound on a type parameter means. "

[Rob Spoor]

And here is the source of that statement, including the exploration.

[Venu Chakravorty]

Thanks Ulrika and Rob.
Sorry for my late reply, my ISP was down.