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。