【Java实战】源码解析为什么覆盖equals方法时总要覆盖hashCode方法

1、背景知识

本文代码基于jdk1.8分析，《Java编程思想》中有如下描述：

另外再看下Object.java对hashCode()方法的说明：

/**     * 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();

对于3点约定翻译如下：

1）在java应用执行期间，只要对象的equals方法的比较操作所用到的信息没有被修改，那么对这同一对象调用多次hashCode方法都必须始终如一地同一个整数。在同一个应用程序的多次执行过程中，每次执行该方法返回的整数可以不一致。

2）如果两个对象根据equals(Object)方法比较是相等的，那么调用这两个对象中任意一个对象的hashCode方法都必须产生同样的整数结果。

3）如果两个对象根据equals(Object)方法比较是不相等的，那么调用这两个对象中任意一个对象的hashCode方法没必要产生不同的整数结果。但是程序猿应该知道，给不同的对象产生截然不同的整数结果，有可能提高散列表（hash table）的性能。

因此，覆盖equals时总是要覆盖hashCode是一种通用的约定，而不是必须的，如果和基于散列的集合（HashMap、HashSet、HashTable）一起工作时，特别是将该对象作为key值的时候，一定要覆盖hashCode，否则会出现错误。那么既然是一种规范，那么作为程序猿的我们就有必要必须执行，以免出现问题。

下面就以HashMap为例分析其必要性

2、HashMap内部实现

常用形式如下：

public class PhoneNumber {    private int areaCode;    private int prefix;    private int lineNumber;    public PhoneNumber(int areaCode, int prefix, int lineNumber) {        this.areaCode = areaCode;        this.prefix = prefix;        this.lineNumber = lineNumber;    }        @Override    public boolean equals(Object o) {        if (this == o) return true;        if (o == null || getClass() != o.getClass()) return false;        PhoneNumber that = (PhoneNumber) o;        if (areaCode != that.areaCode) return false;        if (prefix != that.prefix) return false;        return lineNumber == that.lineNumber;    }    @Override    public int hashCode() {        int result = areaCode;        result = 31 * result + prefix;        result = 31 * result + lineNumber;        return result;    }    public static void main(String[] args){        Map<PhoneNumber,String> phoneNumberStringMap = new HashMap<PhoneNumber,String>();  1）初始化        phoneNumberStringMap.put(new PhoneNumber(123, 456, 7890), "honghailiang");         2）put存储        System.out.println(phoneNumberStringMap.get(new PhoneNumber(123, 456, 7890)));     3）get获取    }}

1）初始化

/**     * Constructs an empty <tt>HashMap</tt> with the default initial capacity     * (16) and the default load factor (0.75).     */    public HashMap() {        this.loadFactor = DEFAULT_LOAD_FACTOR; // all other fields defaulted    }

创建一个具有默认负载因子的HashMap，默认负载因子是0.75

2）put存储

/**     * Associates the specified value with the specified key in this map.     * If the map previously contained a mapping for the key, the old     * value is replaced.     *     * @param key key with which the specified value is to be associated     * @param value value to be associated with the specified key     * @return the previous value associated with <tt>key</tt>, or     *         <tt>null</tt> if there was no mapping for <tt>key</tt>.     *         (A <tt>null</tt> return can also indicate that the map     *         previously associated <tt>null</tt> with <tt>key</tt>.)     */    public V put(K key, V value) {        return putVal(hash(key), key, value, false, true);    }

通过注释可以看出，key值相同的情况下，会将前者覆盖，也就是HashMap中不允许存在重复的Key值。并且该方法是有返回值的，返回key值的上一个value，如果之前没有map则返回null。继续看putVal

/**     * Implements Map.put and related methods     *     * @param hash hash for key     * @param key the key     * @param value the value to put     * @param onlyIfAbsent if true, don't change existing value     * @param evict if false, the table is in creation mode.     * @return previous value, or null if none     */    final V putVal(int hash, K key, V value, boolean onlyIfAbsent,                   boolean evict) {        Node<K,V>[] tab; Node<K,V> p; int n, i;        if ((tab = table) == null || (n = tab.length) == 0)      //tab为空则创建            n = (tab = resize()).length;        if ((p = tab[i = (n - 1) & hash]) == null)               //根据下标获取，如果没有（没发生碰撞（hash值相同））则直接创建            tab[i] = newNode(hash, key, value, null);        else {                                                   //如果发生了碰撞进行如下处理            Node<K,V> e; K k;            if (p.hash == hash &&                ((k = p.key) == key || (key != null && key.equals(k))))                e = p;            else if (p instanceof TreeNode)                      //为红黑数的情况                e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);            else {                                               //为链表的情况，普通Node                for (int binCount = 0; ; ++binCount) {                    if ((e = p.next) == null) {                        p.next = newNode(hash, key, value, null); //链表保存                        if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st                            treeifyBin(tab, hash);                //如果链表长度超过了8则转为红黑树                         break;                    }                    if (e.hash == hash &&                        ((k = e.key) == key || (key != null && key.equals(k))))                        break;                    p = e;                }            }            if (e != null) { // existing mapping for key                     // 写入，并返回oldValue                V oldValue = e.value;                if (!onlyIfAbsent || oldValue == null)                    e.value = value;                afterNodeAccess(e);                return oldValue;            }        }        ++modCount;        if (++size > threshold)          // 超过load factor*current capacity，resize            resize();        afterNodeInsertion(evict);        return null;    }

可以看到第一个参数时key的hash，如下

/**     * Computes key.hashCode() and spreads (XORs) higher bits of hash     * to lower.  Because the table uses power-of-two masking, sets of     * hashes that vary only in bits above the current mask will     * always collide. (Among known examples are sets of Float keys     * holding consecutive whole numbers in small tables.)  So we     * apply a transform that spreads the impact of higher bits     * downward. There is a tradeoff between speed, utility, and     * quality of bit-spreading. Because many common sets of hashes     * are already reasonably distributed (so don't benefit from     * spreading), and because we use trees to handle large sets of     * collisions in bins, we just XOR some shifted bits in the     * cheapest possible way to reduce systematic lossage, as well as     * to incorporate impact of the highest bits that would otherwise     * never be used in index calculations because of table bounds.     */    static final int hash(Object key) {        int h;        return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);    }

综合考虑了速度、作用、质量因素，就是把key的hashCode的高16bit和低16bit异或了一下。因为现在大多数的hashCode的分布已经很不错了，就算是发生了碰撞也用O(logn)的tree去做了。仅仅异或一下，既减少了系统的开销，也不会造成的因为高位没有参与下标的计算(table长度比较小时)，从而引起的碰撞。再回过头来看putVal

1.先判断存有Node数组table是否为null或者大小为0，如果是初始化一个tab并获取它的长度。resize()后面再说，先看下Node的结构

/**     * Basic hash bin node, used for most entries.  (See below for     * TreeNode subclass, and in LinkedHashMap for its Entry subclass.)     */    static class Node<K,V> implements Map.Entry<K,V> {        final int hash;        final K key;        V value;        Node<K,V> next;        Node(int hash, K key, V value, Node<K,V> next) {            this.hash = hash;            this.key = key;            this.value = value;            this.next = next;        }        public final K getKey()        { return key; }        public final V getValue()      { return value; }        public final String toString() { return key + "=" + value; }        public final int hashCode() {            return Objects.hashCode(key) ^ Objects.hashCode(value);        }        public final V setValue(V newValue) {            V oldValue = value;            value = newValue;            return oldValue;        }        public final boolean equals(Object o) {            if (o == this)                return true;            if (o instanceof Map.Entry) {                Map.Entry<?,?> e = (Map.Entry<?,?>)o;                if (Objects.equals(key, e.getKey()) &&                    Objects.equals(value, e.getValue()))                    return true;            }            return false;        }    }

Node实现了链表形式，用于存储hash值没有发生碰撞的hash、key、value，如果发生碰撞则用TreeNode存储，继承自Entry，并最终继承自Node

/**     * Entry for Tree bins. Extends LinkedHashMap.Entry (which in turn     * extends Node) so can be used as extension of either regular or     * linked node.     */    static final class TreeNode<K,V> extends LinkedHashMap.Entry<K,V> {        TreeNode<K,V> parent;  // red-black tree links        TreeNode<K,V> left;        TreeNode<K,V> right;        TreeNode<K,V> prev;    // needed to unlink next upon deletion        boolean red;        TreeNode(int hash, K key, V val, Node<K,V> next) {            super(hash, key, val, next);        }......}

2.以(n - 1) & hash为下标从tab中取出Node，如果不存在，则以hash、Key、value、null为参数new一个Node，存储到以(n - 1) & hash为下标的tab中

3.如果该下标中有值，也就是Node存在。如果为TreeNode，就用putTreeVal进行树节点的存储。否则以链表的形式存储，如果链表长度超过8则转为红黑树存储。

4.如果节点已经存在就替换old value(保证key的唯一性)

5.如果bucket（Node数组）满了(超过load factor*current capacity)，就要resize。

总结：put存储过程：将K/V传给put方法时，它调用hashCode计算hash从而得到Node位置，进一步存储，HashMap会根据当前Node的占用情况自动调整容量(超过Load Facotr则resize为原来的2倍)。可见如果不覆盖hashCode就不能正确的存储。

3）get获取

看完put，再看下get

/**     * Returns the value to which the specified key is mapped,     * or {@code null} if this map contains no mapping for the key.     *     * <p>More formally, if this map contains a mapping from a key     * {@code k} to a value {@code v} such that {@code (key==null ? k==null :     * key.equals(k))}, then this method returns {@code v}; otherwise     * it returns {@code null}.  (There can be at most one such mapping.)     *     * <p>A return value of {@code null} does not <i>necessarily</i>     * indicate that the map contains no mapping for the key; it's also     * possible that the map explicitly maps the key to {@code null}.     * The {@link #containsKey containsKey} operation may be used to     * distinguish these two cases.     *     * @see #put(Object, Object)     */    public V get(Object key) {        Node<K,V> e;        return (e = getNode(hash(key), key)) == null ? null : e.value;    }

get方法又用到了hash()，是根据key的hash和key获取Node，返回的值就是Node的value属性。下面主要看下getNode方法即可

/**     * Implements Map.get and related methods     *     * @param hash hash for key     * @param key the key     * @return the node, or null if none     */    final Node<K,V> getNode(int hash, Object key) {        Node<K,V>[] tab; Node<K,V> first, e; int n; K k;        if ((tab = table) != null && (n = tab.length) > 0 &&            (first = tab[(n - 1) & hash]) != null) {                    //map中存在的情况，不存在则直接返回null            if (first.hash == hash && // always check first node                ((k = first.key) == key || (key != null && key.equals(k))))     //第一个直接命中                return first;            if ((e = first.next) != null) {                             //如果第一个没命中，获取下一个节点                if (first instanceof TreeNode)                    return ((TreeNode<K,V>)first).getTreeNode(hash, key);   //如果下一个节点是TreeNode，则用getTreeNode当时获取                do {                    if (e.hash == hash &&                        ((k = e.key) == key || (key != null && key.equals(k))))    //循环节点链表，直到命中                        return e;                } while ((e = e.next) != null);            }        }        return null;    }

1）第一个直接命中

2）否则，获取下一个节点，如果是红黑树，则从红黑树中获取，否则循环节点链表，直至命中。命中的条件是hash相等且key也相同（基本类型==，自定义类则用equals）。

总结：获取对象时，我们将K传给get，它调用hashCode计算hash从而得到Node位置，并进一步调用==或equals()方法确定键值对。可见为了正确的获取，要覆盖hashCode和equals方法

题外话：当链表长度超过8的时候，java8用红黑树代替了链表，目的是提高性能，这里不展开。HashMap是基于Map接口的实现，存储键值对时，它可以接收null的键值，是非同步的，HashMap存储着Entry(hash, key, value, next)对象。

3、为什么覆盖equals的时候要覆盖hashCode

通过HashMap的实现原理，可以看出当自定义类作为key值存在的时候一定要这样做，但不作为key值可以选择不这样做（但为了规范起见，还是要覆盖，因此就变成了必须的了）。如果将测试代码中的equals或hashCode注释掉都不能得到正确的结果：

public class PhoneNumber {    private int areaCode;    private int prefix;    private int lineNumber;    public PhoneNumber(int areaCode, int prefix, int lineNumber) {        this.areaCode = areaCode;        this.prefix = prefix;        this.lineNumber = lineNumber;    }//    @Override//    public boolean equals(Object o) {//        if (this == o) return true;//        if (o == null || getClass() != o.getClass()) return false;////        PhoneNumber that = (PhoneNumber) o;////        if (areaCode != that.areaCode) return false;//        if (prefix != that.prefix) return false;//        return lineNumber == that.lineNumber;//    }    @Override    public int hashCode() {        int result = areaCode;        result = 31 * result + prefix;        result = 31 * result + lineNumber;        return result;    }    public static void main(String[] args){        Map<PhoneNumber,String> phoneNumberStringMap = new HashMap<PhoneNumber,String>();        phoneNumberStringMap.put(new PhoneNumber(123, 456, 7890), "honghailiang");        System.out.println(phoneNumberStringMap.get(new PhoneNumber(123, 456, 7890)));    }}

上述结果均为null；

题外话Java中的基本类型可以作为key值，包括String类，String类已经覆盖了equals方法和hashCode方法。