please explain Generics compilation error

[raj malhotra]

I have written the following code:

ArrayList<Dvd> ll=new ArrayList<Dvd>();
ll.add(new Dvd());
ll.add(new Dvd());
ll.add(new Dvd());
System.out.println(ll);
Collections.sort(ll);

I am getting compilation error:Bound mismatch: The generic method sort(List<T> ) of type Collections is not applicable for the arguments (ArrayList<Dvd> ) . The inferred type Dvd is not a valid substitute for the bounded parameter <T extends Comparable<? super T>> at Generics1.main.

can anyone explain what Comparable<? super T> means
Thanks
Raj
[ May 29, 2007: Message edited by: raj malhotra ]

[Chandra Bhatt]

Your DVD class must implement Comparable interface so that sort() method could
sort the object of DVD whose references are staying in the ArrayList.



With Regards,

[Chandra Bhatt]

can anyone explain what Comparable<? super T> means

In simple wordings it does mean that it is not must that class T implements
the Comparable interface. If super class of that implements the Comparable
it will also do.

In your case you experiment with what I say:

1- First try by making your Dvd class implement Comparable interface.
2- After that you make a new class and extend the Dvd class from that.
Parent class of Dvd class should implement the Comparable this time so
that you could test the meaning of Comparable<? super T>



Regards,
cmbhatt

[Barry Gaunt]

Following on from what Chandra wrote: class Dvd can implement interface Comparable<Dvd> simply by having a suitable int compareTo(Dvd dvd) method defined.
[ May 29, 2007: Message edited by: Barry Gaunt ]

[raj malhotra]

Got it.Any class ,which implements Comparable interface or whose super class implements it,can be used to sort using Collections.sort(T o)
Thanks Chandra & Barry
[ May 29, 2007: Message edited by: raj malhotra ]