Dog is not abstract and does not override abstract method

[Bud Tippins]

I'm getting the following error when I try to run the code below in NetBeans:
Exception in thread "main" java.lang.ExceptionInInitializerError
at Practice16.main(Practice16.java:13)
Caused by: java.lang.RuntimeException: Uncompilable source code - Dog is not abstract and does not override abstract method compareTo(java.lang.Object) in java.lang.Comparable
at Dog.<clinit>(Practice16.java:3)

Any help in explaining why I am getting this error and how to resolve it would be much appreciated.

Thank you,

import java.util.*;

class Dog implements Comparable {
    int size;
    Dog(int s) { size = s;}
}

public class Practice16 {
  public static void main(String[] args) {
      TreeSet<Integer> i = new TreeSet<Integer>();
      TreeSet<Dog> d = new TreeSet<Dog>();

      d.add(new Dog(1)); d.add(new Dog(2)); d.add(new Dog(1));
      i.add(1); i.add(2); i.add(1);
      System.out.println(d.size() + " " + i.size());
    
  }
}

[Bear Bibeault]

The compiler doesn't like being lied to. You told it that you will implement the java.lang.Comparable interface in Dog, but then you do not.

[Wouter Oet]

The problems is that Dog declares that it implements an interface, Comparable. But it doesn't implement the methods. So the solution is to implement the methods of the Comparable interface.

Small hint: you probably want Comparable<Dog> and when netbeans says that you have compilation error don't run the code.

[Brian Enochson]

Bud,
You are getting this error because any class that implements the interface Comparable has to implement the method

"int compareTo(Object obj)"

;

Your compareTo method will return 1 of 3 values. 0 if the object passed in is equal to yours, 1 if your object is considered greater than the passed in object and -1 if it is less than. What you need to decide what is the criteria for the comparison. In your case, I would use size. Therefore your compareTo method might look like this:

public int compareTo(Object o1) {

        if(!o1 instanceof Dog) {
             // maybe throw an exception.  Shouldn't happen.
             return 1;
        }
        if (this.size == ((Dog) o1).size) {
            return 0;
        }
        else if ((this.size) > ((Dog) o1).size) {
            return 1;
        }
        else {
            return -1;
       }
} 

[Campbell Ritchie]

Good answer, except a tiny disagreement with this bit:

Brian Enochson wrote: . . . Your compareTo method will return 1 of 3 values. 0 if the object passed in is equal to yours, 1 if your object is considered greater than the passed in object and -1 if it is less than. . . .

It returns a negative number if "less than", a positive number if "greater than" and 0 if "the same". 1, -1 and 0 are popular return values, but are not the only permissible values. It is commonly used in code like this

int i = a.compareTo(b);
if (i < 0)
{
   doSomething();
}
else if (i == 0)
{
   doSomethingElse();
}
else // i must be > 0
{
   doSomethingCompletelyDifferent();
}