我也遇到过同样的问题，但我解决了。

//This this your code

private int compareParents(Foo s1, Foo s2) {
    if (s1.getParent() == s2) return -1;
    if (s2.getParent() == s1) return 1;
    return 0;
}

违例是将不同的事物相互比较。

//acceptable
compare between s1.getParent() and s2.getParent()

//acceptable
compare between s1 and s2

//NOT acceptable
compare between s1 and s2.getParent()

//NOT acceptable
compare between s1.getParent() and s2

在我的代码中，我想通过地址的协调来排序。在比较器中，我错误地比较了X和Y，而不是X和X。

//My code:
    private void sortBasedOnX(){
        //addresses is a list of addresses where each address has X and Y
        addresses.sort((o1, o2) -> {

            String a = o1.getAddress().getX(); 
            String b = o2.getAddress().getY(); //<-- this is supposed to be getX

            return Integer.parseInt(a)-Integer.parseInt(b);
        });
    }

//acceptable
compare between o1.getAddress().getX() and o1.getAddress().getX()

//acceptable
compare between o1.getAddress().getY() and o1.getAddress().getY()

//NOT acceptable
compare between o1.getAddress().getX() and o1.getAddress().getY()

//NOT acceptable
compare between o1.getAddress().getX() and o1.getAddress()

//NOT acceptable
compare between o1.getAddress().getX() and o1

从严格意义上讲，Java并不检查一致性，只是在遇到严重问题时才通知您。此外，它也没有从错误中提供太多信息。

我对在我的分类器中发生的事情感到困惑，并做了一个严格的consistencyChecker，也许这将帮助你:

/**
 * @param dailyReports
 * @param comparator
 */
public static <T> void checkConsitency(final List<T> dailyReports, final Comparator<T> comparator) {
  final Map<T, List<T>> objectMapSmallerOnes = new HashMap<T, List<T>>();

  iterateDistinctPairs(dailyReports.iterator(), new IPairIteratorCallback<T>() {
    /**
     * @param o1
     * @param o2
     */
    @Override
    public void pair(T o1, T o2) {
      final int diff = comparator.compare(o1, o2);
      if (diff < Compare.EQUAL) {
        checkConsistency(objectMapSmallerOnes, o1, o2);
        getListSafely(objectMapSmallerOnes, o2).add(o1);
      } else if (Compare.EQUAL < diff) {
        checkConsistency(objectMapSmallerOnes, o2, o1);
        getListSafely(objectMapSmallerOnes, o1).add(o2);
      } else {
        throw new IllegalStateException("Equals not expected?");
      }
    }
  });
}

/**
 * @param objectMapSmallerOnes
 * @param o1
 * @param o2
 */
static <T> void checkConsistency(final Map<T, List<T>> objectMapSmallerOnes, T o1, T o2) {
  final List<T> smallerThan = objectMapSmallerOnes.get(o1);

  if (smallerThan != null) {
    for (final T o : smallerThan) {
      if (o == o2) {
        throw new IllegalStateException(o2 + "  cannot be smaller than " + o1 + " if it's supposed to be vice versa.");
      }
      checkConsistency(objectMapSmallerOnes, o, o2);
    }
  }
}

/**
 * @param keyMapValues 
 * @param key 
 * @param <Key> 
 * @param <Value> 
 * @return List<Value>
 */ 
public static <Key, Value> List<Value> getListSafely(Map<Key, List<Value>> keyMapValues, Key key) {
  List<Value> values = keyMapValues.get(key);

  if (values == null) {
    keyMapValues.put(key, values = new LinkedList<Value>());
  }

  return values;
}

/**
 * @author Oku
 *
 * @param <T>
 */
public interface IPairIteratorCallback<T> {
  /**
   * @param o1
   * @param o2
   */
  void pair(T o1, T o2);
}

/**
 * 
 * Iterates through each distinct unordered pair formed by the elements of a given iterator
 *
 * @param it
 * @param callback
 */
public static <T> void iterateDistinctPairs(final Iterator<T> it, IPairIteratorCallback<T> callback) {
  List<T> list = Convert.toMinimumArrayList(new Iterable<T>() {

    @Override
    public Iterator<T> iterator() {
      return it;
    }

  });

  for (int outerIndex = 0; outerIndex < list.size() - 1; outerIndex++) {
    for (int innerIndex = outerIndex + 1; innerIndex < list.size(); innerIndex++) {
      callback.pair(list.get(outerIndex), list.get(innerIndex));
    }
  }
}

我也遇到过同样的问题，但我解决了。

//This this your code

private int compareParents(Foo s1, Foo s2) {
    if (s1.getParent() == s2) return -1;
    if (s2.getParent() == s1) return 1;
    return 0;
}

违例是将不同的事物相互比较。

//acceptable
compare between s1.getParent() and s2.getParent()

//acceptable
compare between s1 and s2

//NOT acceptable
compare between s1 and s2.getParent()

//NOT acceptable
compare between s1.getParent() and s2

在我的代码中，我想通过地址的协调来排序。在比较器中，我错误地比较了X和Y，而不是X和X。

//My code:
    private void sortBasedOnX(){
        //addresses is a list of addresses where each address has X and Y
        addresses.sort((o1, o2) -> {

            String a = o1.getAddress().getX(); 
            String b = o2.getAddress().getY(); //<-- this is supposed to be getX

            return Integer.parseInt(a)-Integer.parseInt(b);
        });
    }

//acceptable
compare between o1.getAddress().getX() and o1.getAddress().getX()

//acceptable
compare between o1.getAddress().getY() and o1.getAddress().getY()

//NOT acceptable
compare between o1.getAddress().getX() and o1.getAddress().getY()

//NOT acceptable
compare between o1.getAddress().getX() and o1.getAddress()

//NOT acceptable
compare between o1.getAddress().getX() and o1

违反合同通常是指比较者在比较对象时没有提供正确或一致的值。例如，你可能想要执行一个字符串比较，并强制空字符串排序到最后:

if ( one.length() == 0 ) {
    return 1;                   // empty string sorts last
}
if ( two.length() == 0 ) {
    return -1;                  // empty string sorts last                  
}
return one.compareToIgnoreCase( two );

但是这忽略了1和2都为空的情况——在这种情况下，返回了错误的值(1而不是0以显示匹配)，比较器将其报告为违规。它应该写成:

if ( one.length() == 0 ) {
    if ( two.length() == 0 ) {
        return 0;               // BOth empty - so indicate
    }
    return 1;                   // empty string sorts last
}
if ( two.length() == 0 ) {
    return -1;                  // empty string sorts last                  
}
return one.compareToIgnoreCase( two );

在我们的例子中，我们得到这个错误是因为我们不小心颠倒了s1和s2比较的顺序。所以要小心。它显然比下面的要复杂得多，但这是一个例子:

s1 == s2   
    return 0;
s2 > s1 
    return 1;
s1 < s2 
    return -1;

我曾在一段代码中看到过这种情况，其中经常执行空值检查:

if(( A==null ) && ( B==null )
  return +1;//WRONG: two null values should return 0!!!