MET34-J. Follow the general contract when implementing the compareTo method

Objects of a class can be ordered relative to one another. One way to do this is for the class to implement the Comparable interface. Library classes like TreeSet and TreeMap will accept Comparable objects and use their compareTo() methods to sort them. However, a class that implements the compareTo() method in an unexpected way could cause undesirable results. For instance, a TreeSet reporting it does not contain an object that it really does contain, could lead to exploitable behavior.

The general usage contract for compareTo() has been put forth verbatim from the Java SE 6 API [[API 06]]:

The implementpr 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 implementor 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."

In the foregoing description, the notation sgn(expression) designates the mathematical signum function, which is defined to return either -1, 0, or 1 depending on whether the value of the expression is negative, zero or positive.

Do not violate any of these four conditions when implementing the compareTo() method.

Noncompliant Code Example

This noncompliant code example violates the third condition (transitivity) in the contract. This requirement states that the objects that compareTo() considers equal (returns 0) must be ordered the same with respect to other objects. Consider a Card that considers itself equal to any card of the same suit or same rank; otherwise it orders based on rank. This might arise in a game like Uno or Crazy Eights, where you can only place a card on the pile that shares a suit or rank with the top card on the pile.

public final class Card implements Comparable {
  private String suit;
  private int rank;

  public Card(String s, int r) {
    if (s == null)
      throw new NullPointerException();
    suit = s;
    rank = r;
  }

  public boolean equals(Object o) {
    if (o instanceof Card){
      Card c = (Card)o;
      return suit.equals(c.suit) || (rank == c.rank); // bad
    }
    return false;
  }

//This method violates its contract
  public int compareTo(Object o){
    if (o instanceof Card){
      Card c = (Card)o;
      if(suit.equals(c.suit)) 
        return 0;
      if((c.rank >= rank + Integer.MIN_VALUE) && (c.rank <= rank + Integer.MAX_VALUE)) // check for integer underflow/overflow
        return c.rank - rank;
    }
    throw new ClassCastException();
  }

  public static void main(String[] args) {
    Card a = new Card("Clubs", 2);
    Card b = new Card("Clubs", 10);
    Card c = new Card("Hearts", 7);
    System.out.println(a.compareTo(b)); //returns 0
    System.out.println(a.compareTo(c)); //returns a negative number
    System.out.println(b.compareTo(c)); //returns a positive number
  }
}

Here, the comparison between (a,c) yields that c is larger. But, (b,c) yields b as larger. This means b must be larger than a. However, (a,b) results in the value 0 (same).

Compliant Solution

Make sure you fulfill the compareTo() contract, and don't forget to make sure your corresponding equals() method matches with compareTo().

public final class Card implements Comparable{
  private String suit;
  private int rank;

  public Card(String s, int r) {
    if (s == null)
      throw new NullPointerException();
    suit = s;
    rank = r;
  }

  public boolean equals(Object o) {
    if (o instanceof Card){
      Card c=(Card)o;
      return suit.equals(c.suit) && (rank == c.rank); // good
    }
    return false;
  }

  //this method fulfills its contract
  public int compareTo(Object o){
    if (o instanceof Card){
      Card c=(Card)o;
      if(suit.equals(c.suit)) 
        return c.rank - rank;
      return suit.compareTo(c.suit);
    }
    throw new ClassCastException();
  }

  public static void main(String[] args) {
    Card a = new Card("Clubs", 2);
    Card b = new Card("Clubs", 10);
    Card c = new Card("Hearts", 7);
    System.out.println(a.compareTo(b)); //returns 0
    System.out.println(a.compareTo(c)); //returns a negative number
    System.out.println(b.compareTo(c)); //returns a negative number
  }
}

As per the ordering, c is larger than both a and b and the comparison (a,b) produces an "equal" result. This maintains the compareTo contract.

Exceptions

MET34-EX1: In some situations it may be necessary to violate the condition that (x.compareTo(y) == 0) == (x.equals(y) ). However, as this will make an object Comparable, it should not be used in most data structures (like TreeSet and TreeMap) that deal with Comparable objects. It would be better to use a separate method entirely. If compatibility issues demand the use of this non-standard compareTo() method, care should be taken through comments or other means that the object is not used in data structures that expect a standard Comparable.

Risk Assessment

Violating the general contract when implementing the compareTo() method can lead to unexpected results, possibly leading to invalid comparisons and information disclosure.

Automated Detection

TODO

Related Vulnerabilities

Search for vulnerabilities resulting from the violation of this rule on the CERT website.

Other Languages

This rule appears in the C++ Secure Coding Standard as ARR40-CPP. Use a Valid Ordering Rule.

References

[[API 06]] method compareTo()
[[JLS 05]]