为什么系列之重写equals方法必须重写hasCode方法?
Object源代码及注释
equals是Object的公有方法,那么我们通常都会在自己的类中重写这个equals方法,同时必须重写hasCode方法,知道为什么重写equals方法必须重写hasCode方法呢?
/**
* Returns a hash code value for the object. This method is
* supported for the benefit of hash tables such as those provided by
* {@link java.util.HashMap}.
* <p>
* The general contract of {@code hashCode} is:
* <ul>
* <li>Whenever it is invoked on the same object more than once during
* an execution of a Java application, the {@code hashCode} method
* must consistently return the same integer, provided no information
* used in {@code equals} comparisons on the object is modified.
* This integer need not remain consistent from one execution of an
* application to another execution of the same application.
* <li>If two objects are equal according to the {@code equals(Object)}
* method, then calling the {@code hashCode} method on each of
* the two objects must produce the same integer result.
* <li>It is <em>not</em> required that if two objects are unequal
* according to the {@link java.lang.Object#equals(java.lang.Object)}
* method, then calling the {@code hashCode} method on each of the
* two objects must produce distinct integer results. However, the
* programmer should be aware that producing distinct integer results
* for unequal objects may improve the performance of hash tables.
* </ul>
* <p>
* As much as is reasonably practical, the hashCode method defined by
* class {@code Object} does return distinct integers for distinct
* objects. (This is typically implemented by converting the internal
* address of the object into an integer, but this implementation
* technique is not required by the
* Java™ programming language.)
*
* @return a hash code value for this object.
* @see java.lang.Object#equals(java.lang.Object)
* @see java.lang.System#identityHashCode
*/
public native int hashCode();
/**
* Indicates whether some other object is "equal to" this one.
* <p>
* The {@code equals} method implements an equivalence relation
* on non-null object references:
* <ul>
* <li>It is <i>reflexive</i>: for any non-null reference value
* {@code x}, {@code x.equals(x)} should return
* {@code true}.
* <li>It is <i>symmetric</i>: for any non-null reference values
* {@code x} and {@code y}, {@code x.equals(y)}
* should return {@code true} if and only if
* {@code y.equals(x)} returns {@code true}.
* <li>It is <i>transitive</i>: for any non-null reference values
* {@code x}, {@code y}, and {@code z}, if
* {@code x.equals(y)} returns {@code true} and
* {@code y.equals(z)} returns {@code true}, then
* {@code x.equals(z)} should return {@code true}.
* <li>It is <i>consistent</i>: for any non-null reference values
* {@code x} and {@code y}, multiple invocations of
* {@code x.equals(y)} consistently return {@code true}
* or consistently return {@code false}, provided no
* information used in {@code equals} comparisons on the
* objects is modified.
* <li>For any non-null reference value {@code x},
* {@code x.equals(null)} should return {@code false}.
* </ul>
* <p>
* The {@code equals} method for class {@code Object} implements
* the most discriminating possible equivalence relation on objects;
* that is, for any non-null reference values {@code x} and
* {@code y}, this method returns {@code true} if and only
* if {@code x} and {@code y} refer to the same object
* ({@code x == y} has the value {@code true}).
* <p>
* Note that it is generally necessary to override the {@code hashCode}
* method whenever this method is overridden, so as to maintain the
* general contract for the {@code hashCode} method, which states
* that equal objects must have equal hash codes.
*
* @param obj the reference object with which to compare.
* @return {@code true} if this object is the same as the obj
* argument; {@code false} otherwise.
* @see #hashCode()
* @see java.util.HashMap
*/
public boolean equals(Object obj) {
return (this == obj);
}
上面是Object对象中hasCode和equals的签名和方法说明。
为啥重写equals方法
判断一些对象是否等于当前对象,实现在非空对象映射的等价关系。this==obj,由此可知,这是一个引用对象的等价判断,只有在this和obj是指向用一个应用时才返回true.这是判定实例的唯一性,我们通常使用equals时,其实并不想判断实例的唯一性,而是想要比较里面的几个(自定义)属性值是否相等,来判断是否为同一个"东西"。
契约:
reflexive(自反性):x为非null时,x.equals(x),一定返回true.
symmetric(对称性):x,y都不为null,如果x.equals(y)返回true,那么y.equals(x).
transitive(传递性):x,y,z都不为null,如果x.equals(y),y.equals(z)返回true,那么x.equals(z)返回true.
consistent(一致性):对于任何非null的x\y,在没有修改这些对象信息的前提下,多次调用下,x.equals(y)结果是一致的。
非null的x,x.equals(null)返回false.
下面是Book类的equals重写实现,通过bookN、bookName确定是否为同一本书(东西),而不是判断两个对象的引用是否为同一个。
@Override
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
Book book = (Book) o;
return Objects.equal(bookNo, book.bookNo) &&
Objects.equal(bookName, book.bookName);
}
为啥重写hasCode方法
hasCode锲约:
1.任何时候在没有修改对象的前提下,多次调用同一个对象的hasCode方法,这个返回值必须是同一个Integer值。
2.如果两个对象equals结果为true,那么hasCode的值必须相同。
3.如果两个对象equals结果为false,那么hasCode就不一定不相同。
如果重写equals方法时,不去重写hasCode方法,那么必然会违背第二条锲约。所以每次hasCode必须跟随equals重写而重写。
扩展知识
是不是有人就是问:如果就不重写hasCode,不行吗?就违背锲约不可以吗?只equals重写有什么问题吗?
那么这里就会引申出hasCode在集合中的应用场景。
Java中的集合(Collection)有两类,一类是List,再有一类是Set。前者集合内的元素是有序的,元素可以重复;后者元素无序,但元素不可重复。那么这里就有一个比较严重的问题了:要想保证元素不重复,可两个元素是否重复应该依据什么来判断呢?这就是 Object.equals方法了。但是,如果每增加一个元素就检查一次,那么当元素很多时,后添加到集合中的元素比较的次数就非常多了。也就是说,如果集合中现在已经有1000个元素,那么第1001个元素加入集合时,它就要调用1000次equals方法。这显然会大大降低效率。
于是,Java采用了哈希表的原理。哈希算法也称为散列算法,是将数据依特定算法直接指定到一个地址上。可以这样简单理解,hashCode方法实际上返回的就是对象存储位置的映像。
当集合中添加元素时,先调用hasCode方法,如果这个位置没有元素,那么就直接存储在这里,如果有元素,那么再调用equals方法与新元素比较,如果相同不存了,如果不相同则说明发生碰撞(冲突),碰撞解决方法多样,最终都会存储在一个合适的位置。有了hasCode后,调用equals的次数大大降低,一般一两次就搞定了。
扩展的内容有兴趣可以自行深入学习。为什么系列采用简短,易懂方式归纳总结一些工作中、面试中常常出现的问题进行解答。如果有错误请及时联系我,以免误导他人。