Re: Generics compiler warning

From:

"Oliver Wong" <owong@castortech.com>

Newsgroups:

comp.lang.java.programmer

Date:

Thu, 27 Apr 2006 17:15:51 GMT

Message-ID:

<b%64g.840$nq3.675@clgrps12>

"Christian Pontesegger" <christian.pontesegger@web.de> wrote in message
news:4450f1ab$0$578$79720d31@newsreader.inode.at...

Hi,

I was trying to create a generic compare method for objects of different
type. If both objects are of the same type and implement Comparable then
use the compareTo method. Else use their toString method and compare their
results.

My code looks like this:

private int compare(Object o1, Object o2) {
  if (o1 == null)
    return -1;
  if (o2 == null)
    return 11;

  if ((o1.getClass().equals(o2.getClass())) && (o1 instanceof
       Comparable))
            return ((Comparable) o1).compareTo((Comparable) o2);

  // compare anything else converted to string
  return o1.toString().compareTo(o2.toString());
}

It works as expected, but I get this annoying compiler warning

"Type safety: The method compareTo(Object) belongs to the raw type
Comparable. References to generic type Comparable<T> should be
parameterized"

How should I code my method in a clean way?

    Perhaps you could use method overloading, one for where the two
arguments implement Comparable, and one for where only one, or none of them,
implement Comparable.

<code>
  private <T extends Comparable<T>> int compare(T o1, T o2) {
    if (o1 == null) {
      return -1;
    }
    if (o2 == null) {
      return 11;
    }

    if (o1.getClass().equals(o2.getClass())) {
      return ((T) o1).compareTo((T) o2);
    }
    return compare((Object) o1, (Object) o2);
  }

  private int compare(Object o1, Object o2) {
    if (o1 == null) {
      return -1;
    }
    if (o2 == null) {
      return 11;
    }
    // compare anything else converted to string
    return o1.toString().compareTo(o2.toString());
  }
</code>

    - Oliver