java compareTo() for rawtype Comparable

Question

 public class Test3{


public static void main(String[] args)
{
    String s="1";
    String s1="2";
    System.out.println(Compare.compare(s, s1));


}
}

class Compare {
public static Comparable compare ( Comparable o1, Comparable o2 ) {
if (o1.compareTo(o2) > 0)
return o1;
else
return o2;
}
}

The above code output 2, which is correct if String.compareTo(String str) is used. However the type of o1 is actually raw type Comparable not Comparable<String> and even if it is Comparable<String>, Comparable o2 cannot be passed as parameter.

So, what o1.compareTo(o2) actually call? and How it actually work

Answer 1

This is a really unclear question to begin with. Your first example uses raw types, so Comparable#compareTo expects a certain type that T (but because you are using raw types, this is an Object) that it should compare against, well a raw Comparable (o2 in your case) fits that case (it is an Object after all).

When you change your class to be:

static class Compare {
    public Comparable<String> compare(Comparable<String> left, Comparable<String> right) {
        if (left.compareTo(right) > 0) {
            return left;
        } else {
            return right;
        }
    }
}

This does not work because compareTo expects a type T (String actually), but you are passing Comparable<String>. This is actually really easy to understand once you look at the Comparable interface in the jdk sources...

A correct implementation using generics IMO would be:

static class Compare<T extends Comparable<? super T>> {
    public T compare(T left, T right) {
        if (left.compareTo(right) > 0) {
            return left;
        } else {
            return right;
        }
    }
}