Question

When implementing compareTo(), does the degree of "difference" need to be taken into account?

For instance, if I have 3 objects, C1, C2, and C3, such that C1 < C2 < C3.

Should C1.compareTo(C2) return an integer that is less than C2.compareTo(C3)?

The documentation for the Comparable interface doesn’t seem to specify one way or another, so I’m guessing the degree doesn’t matter, but it would be nice to know if there is some advantage returning a specific number (for example, improving TreeSet sort speed or something).

http://java.sun.com/j2se/1.5.0/docs/api/java/lang/Comparable.html#compareTo(T)

Answer 1

Interesting question, but nonetheless no, the magnitude of the int has no significance as per Comparable<T> and Comparator<T> specifications, only the sign. Conceivably some sorting algorithm can additionally specify that they can take “hints” from the magnitude, but I’m not sure how practical that would be for comparison-based sorting, since we really need only to know if a < b, a == b, or a > b (which is really what Comparable and Comparator are OOP abstractions of).

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Now it needs to be said that there may be a hidden intention here of using the subtraction idiom for comparing numeric values, i.e. something like this:

public int compare(T t1, T t2) {
    return t1.intField - t2.intField;
}

Do note that this comparison method is potentially broken, due to possible overflow when the difference between the two numbers is greater than Integer.MAX_VALUE. In fact, this is one of the puzzles covered in Java Puzzlers.

To demonstrate, consider the following snippet (taken from the book):

int x = -2000000000;
int z = 2000000000;
System.out.println(x - z); // prints a positive number due to overflow

Clearly x < z, and yet x - z is a positive number. Beware of using this subtraction idiom: it’s always much safer to do an explicit comparison and return -1, 0, or 1 instead.