Source code analysis of comparable interface in jdk8
/**
- Created by caoxiaohong on 17/11/18 23:14.
*/
import java.util.*;
/**
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This interface imposes a total ordering on the objects of each class that
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implements it. This ordering is referred to as the class's natural
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ordering,and the class's compareTo method is referred to as
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its natural comparison method.
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这一接口会对实现了它的类施加一个整体的顺序.这一顺序被认为是类的自然顺序,类的比较方法compareTo()也被认为是自然比较方法.
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Lists (and arrays) of objects that implement this interface can be sorted
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automatically by {@link Collections#sort(List) Collections.sort} (and
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{@link Arrays#sort(Object[]) Arrays.sort}). Objects that implement this
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interface can be used as keys in a {@linkplain SortedMap sorted map} or as
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elements in a {@linkplain SortedSet sorted set},without the need to
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specify a {@linkplain Comparator comparator}.
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实现类这一接口的List类对象使用Collections.sort方法实现自动排序(升序),数组使用Arrays.sort()方法实现升序排序.实现这一接口的对象在
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有序Map中,有序是按照key进行排序的;在有序Set中,是按照set集合中的元素排序的.而使用这些方法时,我们并不需要指定比较器comparator(说明:
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这些排序都是默认升序排序,且排序字段只有一个.如果一个类有多个排序字段,要对这个类集合进行排序,则需要重写比较器方法).
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The natural ordering for a class C is said to be consistent
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with equals if and only if e1.compareTo(e2) == 0 has
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the same boolean value as e1.equals(e2) for every
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e1 and e2 of class C. Note that null
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is not an instance of any class,and e.compareTo(null) should
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throw a NullPointerException even though e.equals(null)
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returns false.
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对于类C的任意变量e1和e2,当且仅当e1.compareTo(e2) == 0的和e1.equals(e2)有相同的返回值时,类的自然排序才能被认为是和equals方法的
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结果保持一致的.
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注意:虽然e.equals(null)返回值为false,但是null不是任何类的实例,所以如果调用方法e.compareTo(null)应该抛出异常NullPointerException
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It is strongly recommended (though not required) that natural orderings be
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consistent with equals. This is so because sorted sets (and sorted maps)
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without explicit comparators behave "strangely" when they are used with
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elements (or keys) whose natural ordering is inconsistent with equals. In
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particular,such a sorted set (or sorted map) violates the general contract
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for set (or map),which is defined in terms of the equals
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method.
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我们强烈建议(尽管并不是必须的):自然排序应该和equals结果保持一致(这是因为自然排序用到了compare方法,这里的意思是需要满足关系:
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e1.compareTo(e2) == 0的和e1.equals(e2)有相同的返回值).这是因为没有明确比较器的有序set(和有序map)
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(什么叫没有明确比较器?对于TreeSet和TreeMap,都有多个实例构造函数,而其中有一个无参构造函数,就指定了比较器comparator = null;同时,* 这也说明了,如果你想在建立有序set或者有序map时就指定它的排序方法,那么可以给构造函数传入一个比较器参数即可.),如果自然排序不能和equals方法
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保持一致,那么它们会表现出一些诡异的行为.而且,这样的有序set(或者map)和equals中通用规范是矛盾的。
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For example,if one adds two keys a and b such that
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(!a.equals(b) && a.compareTo(b) == 0) to a sorted
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set that does not use an explicit comparator,the second add
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operation returns false (and the size of the sorted set does not increase)
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because a and b are equivalent from the sorted set's
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perspective.
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举个例子:如果向一个没有明确比较器的有序set中添加2个值a和b(a.equals(b)值为false,而 a.compareTo(b) == 0 值为true(a==b)),* 那么第二次的add操作会失败,因为从有序set的角度看,a和b是等值的.(出现这种事情就很诡异了,明明a在add之后,b再add时,这是两个不同的值,应该被正常
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添加到集合中,但是却被拒绝了,因为add时,使用到了方法compare,去比较插入的值是否存在,而根据返回结果为0,这样二者就被认为是相同的值.所以
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我们一再强调:为避免这种异常,自然排序要和equals结果保持一致,必须满足e1.compareTo(e2) == 0的和e1.equals(e2)始终有相同的返回值)
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Virtually all Java core classes that implement Comparable have natural
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orderings that are consistent with equals. One exception is
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java.math.BigDecimal,whose natural ordering equates
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BigDecimal objects with equal values and different precisions
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(such as 4.0 and 4.00).
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实质上,所有实现了Comparable接口的java核心类,都满足自然排序的要求.唯一的例外类是:BigDecimal类.它的自然排序要求是:值相等而精度是不等的.
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所以,精度不同但值相同的两个BigDecimal对象,它们的equals方法返回值应该为true,而compare()方法应该返回0:
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测试用例如下:
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import java.math.BigDecimal;
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public class testCode {
public static void main(String[] args){
BigDecimal a=new BigDecimal(2.30);
BigDecimal b=new BigDecimal(2.3);
System.out.println("a.equals(b): "+a.equals(b));
System.out.println("a.compareTo(b): "+a.compareTo(b));
}
}
输出结果:
a.equals(b): true
a.compareTo(b): 0
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For the mathematically inclined,the relation that defines
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the natural ordering on a given class C is:
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{(x,y) such that x.compareTo(y) <= 0}.
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