Help in generics....

[Nitish Bangera]

   class Vehicle { }
class Car extends Vehicle { }
class Sedan extends Car { }
class Garage<V> {
private V v;
public V get() { return this.v; }
public void put(V v) { this.v = v; }
}
public class GarageAdmin {
private Object object = new Object();
private Vehicle vehicle = new Vehicle();
private Car car = new Car();
private Sedan sedan = new Sedan();
public void doE(Garage<? super Car> g) {
g.put(object); // (1)
g.put(vehicle); // (2)
g.put(car); // (3)
g.put(sedan); // (4)
object = g.get(); // (5)
vehicle = g.get(); //6
car = g.get(); //7
sedan = g.get(); //8
}
}
Select the two correct answers.
(a) The call to the put() method in statements (1) - (2) will compile.
(b) The call to the put() method in statements (3) - (4) will compile.
(c) The assignment statement (5) will compile.
(d) The assignment statements (6), (7), and (8) will compile.
 

This question is from SCJP 6 khalid mughal. The answer to this question is b and c.

well i understand that car can be added but how is sedan put because it says car or super of car(Vehicle or Object).....

Also please explain the other <? extends Car> also.

[Muhammad Khojaye]

For Garage<? super Car> we can pass either Car or any subclass of the Car to the add() method, but we cannot pass a super class of Car. Also see this

[Nitish Bangera]

I guess i confused myself..... This example is like Garage<? super Car> g = new Garage<Car/SUpertype>();

I confused it to to the example given in K&B page 619 where they pass List<Animal> a. reference "a" is passed to a method(List<? super Dog> e) But still a little confusion..... why does it allow as we pass a list<Animal> to a method(list<? super Dog>.)

with raw type object references its not possible as the compiler has no idea which instance it points to or which super type it is going to point to.

I guess i understand a little bit over the problem here..... It is because the V in garage cannot be determined at compile time where as in animal and dog it was determined at compile time that the super of dog is animal.

[Nitish Bangera]

I really confused myself big time...but i am clear now. When we pass a List<animal> a to a method(List<? super Dog> e)..here we are dealing with references and the compiler knows it. Well if we are talking about List<? super Dog> a = something. compiler really doesn't know what the something object will be that's why we cannot add anything into the list.