right way to implement compareTo() method?

[Anthony Karta]

the method below is sorting (descending) the student objects by their id (integer type).

I am not sure if I implemented it correctly (good design? etc).

can we return any value as long as they are positive, negative and zero? how the Collections.sort() implementation's codes use/understand those values?

    
    public int compareTo(StudentComparable sc) {
        int returnValue = 0;
        if (this.id > sc.id)
            returnValue = -1;     // any negative number?
        if (this.id < sc.id)
            returnValue = 1;      // any positive number?
        if (this.id == sc.id)
            returnValue = 0;
 
        return returnValue;
    }

[Christophe Verré]

Any number will do. The calling method does not look at the value, but at the sign, positive, negative or zero.

[Naseem Khan]

-1 means left less than right

0 means both equal

+1 means left greater than right

Your code should be like this..

public int compareTo(StudentComparable sc) {        
  int returnValue = 0;        
  if (this.id < sc.id){            
    returnValue = -1;   
  }
  else if(this.id > sc.id){
    returnValue = 1;     
  }
  else{
    returnValue = 0;
  }
  return returnValue;    
}

From javadoc...

compareTo

public int compareTo(Object o)

Compares this object with the specified object for order. Returns a negative integer, zero, or a positive integer as this object is less than, equal to, or greater than the specified object.

In the foregoing description, the notation sgn(expression) designates the mathematical signum function, which is defined to return one of -1, 0, or 1 according to whether the value of expression is negative, zero or positive. The implementor must ensure sgn(x.compareTo(y)) == -sgn(y.compareTo(x)) for all x and y. (This implies that x.compareTo(y) must throw an exception iff y.compareTo(x) throws an exception.)

The implementor must also ensure that the relation is transitive: (x.compareTo(y)>0 && y.compareTo(z)>0) implies x.compareTo(z)>0.

Finally, the implementer must ensure that x.compareTo(y)==0 implies that sgn(x.compareTo(z)) == sgn(y.compareTo(z)), for all z.

It is strongly recommended, but not strictly required that (x.compareTo(y)==0) == (x.equals(y)). Generally speaking, any class that implements the Comparable interface and violates this condition should clearly indicate this fact. The recommended language is "Note: this class has a natural ordering that is inconsistent with equals."


Parameters:
o - the Object to be compared.
Returns:
a negative integer, zero, or a positive integer as this object is less than, equal to, or greater than the specified object.

Naseem
[ July 23, 2006: Message edited by: Naseem Khan ]