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
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
}
}/**
* 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
/**
* 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);
}/**
* 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。
/**
* 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;
}/**
* 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;
}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；【Java实战】源码解析为什么覆盖equals方法时总要覆盖hashCode方法
原文：http://blog.csdn.net/honghailiang888/article/details/71711454