/*
 * @(#)Comparable.java	1.9 98/09/30
 *
 * Copyright 1997, 1998 by Sun Microsystems, Inc.,
 * 901 San Antonio Road, Palo Alto, California, 94303, U.S.A.
 * All rights reserved.
 * 
 * This software is the confidential and proprietary information
 * of Sun Microsystems, Inc. ("Confidential Information").  You
 * shall not disclose such Confidential Information and shall use
 * it only in accordance with the terms of the license agreement
 * you entered into with Sun.
 */

// IBM JDK is based on JDK1.2 so I rent this JDK1.2 source. YAMAUCHI
// package java.lang;
package jp.ac.tohoku.is.archi.mpi2.jdk12src;

/**
 * This interface imposes a total ordering on the objects of each class that
 * implements it.  This ordering is referred to as the class's <i>natural
 * ordering</i>, and the class's <tt>compareTo</tt> method is referred to as
 * its <i>natural comparison method</i>.<p>
 *
 * Lists (and arrays) of objects that implement this interface can be sorted
 * automatically by <tt>Collections.sort</tt> (and <tt>Arrays.sort</tt>).
 * Objects that implement this interface can be used as keys in a sorted map
 * or elements in a sorted set, without the need to specify a comparator.<p>
 *
 * A class's natural ordering is said to be <i>consistent with equals</i> if
 * and only if <tt>(e1.compareTo((Object)e2)==0)</tt> has the same boolean
 * value as <tt>e1.equals((Object)e2)</tt> for every <tt>e1</tt> and
 * <tt>e2</tt> of class <tt>C</tt>.<p>
 *
 * It is strongly recommended (though not required) that natural orderings be
 * consistent with equals.  This is so because sorted sets (and sorted maps)
 * without explicit comparators behave "strangely" when they are used with
 * elements (or keys) whose natural ordering is inconsistent with equals.  In
 * particular, such a sorted set (or sorted map) violates the general contract
 * for set (or map), which is defined in terms of the <tt>equals</tt>
 * operation.<p>
 *
 * For example, if one adds two keys <tt>a</tt> and <tt>b</tt> such that
 * <tt>(a.equals((Object)b) && a.compareTo((Object)b) != 0)</tt> to a sorted
 * set that does not use an explicit comparator, the second <tt>add</tt>
 * operation returns false (and the size of the sorted set does not increase)
 * because <tt>a</tt> and <tt>b</tt> are equivalent from the sorted set's
 * perspective.<p>
 *
 * Virtually all Java core classes that implement comparable have natural
 * orderings that are consistent with equals.  One exception is
 * <tt>java.math.BigDecimal</tt>, whose natural ordering equates
 * <tt>BigDecimals</tt> with equal values and different precisions (such as
 * 4.0 and 4.00).<p>
 *
 * For the mathematically inclined, the <i>relation</i> that defines
 * the natural ordering on a given class C is:<pre>
 *       {(x, y) such that x.compareTo((Object)y) &lt;= 0}.
 * </pre> The <i>quotient</i> for this total order is: <pre>
 *       {(x, y) such that x.compareTo((Object)y) == 0}.
 * </pre>
 *
 * It follows immediately from the contract for <tt>compareTo</tt> that the
 * quotient is an <i>equivalence relation</i> on <tt>C</tt>, and that the
 * natural ordering is a <i>total order</i> on <tt>C</tt>.  When we say that a
 * class's natural ordering is <i>consistent with equals</i>, we mean that the
 * quotient for the natural ordering is the equivalence relation defined by
 * the class's <tt>equals(Object)</tt> method:<pre>
 *     {(x, y) such that x.equals((Object)y)}.
 * </pre>
 *
 * @author  Josh Bloch
 * @version 1.9 09/30/98
 * @see java.util.Comparator
 * @see java.util.Collections#sort(java.util.List)
 * @see java.util.Arrays#sort(Object[])
 * @see java.util.SortedSet
 * @see java.util.SortedMap
 * @see java.util.TreeSet
 * @see java.util.TreeMap
 * @since JDK1.2
 */

public interface Comparable {
    /**
     * 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.<p>
     *
     * The implementor must ensure <tt>sgn(x.compareTo(y)) ==
     * -sgn(y.compareTo(x))</tt> for all <tt>x</tt> and <tt>y</tt>.  (This
     * implies that <tt>x.compareTo(y)</tt> must throw an exception iff
     * <tt>y.compareTo(x)</tt> throws an exception.)<p>
     *
     * The implementor must also ensure that the relation is transitive:
     * <tt>(x.compareTo(y)&gt;0 &amp;&amp; y.compareTo(z)&gt;0)</tt> implies
     * <tt>x.compareTo(z)&gt;0</tt>.<p>
     *
     * Finally, the implementer must ensure that <tt>x.compareTo(y)==0</tt>
     * implies that <tt>sgn(x.compareTo(z)) == sgn(y.compareTo(z))</tt>, for
     * all <tt>z</tt>.<p>
     *
     * It is strongly recommended, but <i>not</i> strictly required that
     * <tt>(x.compareTo(y)==0) == (x.equals(y))</tt>.  Generally speaking, any
     * class that implements the <tt>Comparable</tt> 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."
     * 
     * @param   o the Object to be compared.
     * @return  a negative integer, zero, or a positive integer as this object
     *		is less than, equal to, or greater than the specified object.
     * 
     * @throws ClassCastException if the specified object's type prevents it
     *         from being compared to this Object.
     */
    public int compareTo(Object o);
}
