java.util
public interface Comparator<T>
The ordering imposed by a Comparator c on a set of elements S is said to be consistent with equals if and only if (compare((Object)e1, (Object)e2)==0) has the same boolean value as e1.equals((Object)e2) for every e1 and e2 in S.
Caution should be exercised when using a comparator capable of imposing an ordering inconsistent with equals to order a sorted set (or sorted map). Suppose a sorted set (or sorted map) with an explicit Comparator c is used with elements (or keys) drawn from a set S. If the ordering imposed by c on S is inconsistent with equals, the sorted set (or sorted map) will behave "strangely." In particular the sorted set (or sorted map) will violate the general contract for set (or map), which is defined in terms of equals.
For example, if one adds two keys a and b such that (a.equals((Object)b) && c.compare((Object)a, (Object)b) != 0) to a sorted set with comparator c, the second add operation will return false (and the size of the sorted set will not increase) because a and b are equivalent from the sorted set's perspective.
Note: It is generally a good idea for comparators to implement java.io.Serializable, as they may be used as ordering methods in serializable data structures (like TreeSet, TreeMap). In order for the data structure to serialize successfully, the comparator (if provided) must implement Serializable.
For the mathematically inclined, the relation that defines the total order that a given comparator c imposes on a given set of objects S is:
{(x, y) such that c.compare((Object)x, (Object)y) <= 0}.
The quotient for this total order is:
{(x, y) such that c.compare((Object)x, (Object)y) == 0}.
It follows immediately from the contract for compare that the
quotient is an equivalence relation on S, and that the
natural ordering is a total order on S. When we say that
the ordering imposed by c on S is consistent with
equals, we mean that the quotient for the natural ordering is the
equivalence relation defined by the objects' equals(Object)
method(s):
{(x, y) such that x.equals((Object)y)}.
This interface is a member of the Java Collections Framework.
Since: 1.2
Version: 1.21, 06/22/03
See Also: Comparable Arrays TreeMap TreeSet SortedMap SortedSet java.io.Serializable
| Method Summary | |
|---|---|
| int | compare(T o1, T o2)
Compares its two arguments for order. |
| boolean | equals(Object obj)
Indicates whether some other object is "equal to" this
Comparator. |
The implementor must ensure that sgn(compare(x, y)) == -sgn(compare(y, x)) for all x and y. (This implies that compare(x, y) must throw an exception if and only if compare(y, x) throws an exception.)
The implementor must also ensure that the relation is transitive: ((compare(x, y)>0) && (compare(y, z)>0)) implies compare(x, z)>0.
Finally, the implementer must ensure that compare(x, y)==0 implies that sgn(compare(x, z))==sgn(compare(y, z)) for all z.
It is generally the case, but not strictly required that (compare(x, y)==0) == (x.equals(y)). Generally speaking, any comparator that violates this condition should clearly indicate this fact. The recommended language is "Note: this comparator imposes orderings that are inconsistent with equals."
Parameters: o1 the first object to be compared. o2 the second object to be compared.
Returns: a negative integer, zero, or a positive integer as the first argument is less than, equal to, or greater than the second.
Throws: ClassCastException if the arguments' types prevent them from being compared by this Comparator.
comp1.equals(comp2) implies that sgn(comp1.compare(o1,
o2))==sgn(comp2.compare(o1, o2)) for every object reference
o1 and o2.Note that it is always safe not to override Object.equals(Object). However, overriding this method may, in some cases, improve performance by allowing programs to determine that two distinct Comparators impose the same order.
Parameters: obj the reference object with which to compare.
Returns: true only if the specified object is also
a comparator and it imposes the same ordering as this
comparator.