HashMap 源码解析
- 2023-09-02  广东
- 本文字数:4580 字 - 阅读完需:约 15 分钟 
大纲
- 基础入门 
- 数组的优势/劣势 
- 链表的优势/劣势 
- 有没有一种方式整合 2 种数据结构的优势?散列表 
- 散列表的特点? 
- 什么是哈希? 
- HashMap 原理讲解 
- HashMap 的继承体系是怎样的? 
- Node 的数据结构分析? 
- 底层存储结构分析? 
- put 数据原理分析? 
- 什么是 hash 碰撞? 
- 什么是链化? 
- jdk8 为什么引入红黑树? 
- HashMap 扩容原理? 
- 手撕源码 
核心的概念
核心的属性
默认的初始化容量:static final int DEFAULT_INITIAL_CAPACITY = 1 << 4
最大容量:static final int MAXIMUM_CAPACITY = 1 << 30
默认的负载因子:static final float DEFAULT_LOAD_FACTOR = 0.75f
树化阈值:static final int TREEIFY_THRESHOLD = 8
树退化阈值:static final int UNTREEIFY_THRESHOLD = 6
可以被树化的散列表最小容量:static final int MIN_TREEIFY_CAPACITY = 64
散列表:transient Node<K,V>[] table 第一次使用才初始化
键值对的数量:transient int size
扩容阈值:int threshold
负载因子:final float loadFactor
Node
散列表节点的基本类型
static class Node<K,V> implements Map.Entry<K,V> {        final int hash;        final K key;        V value;        Node<K,V> next;}
构造器
HashMap 提供了多个构造器,最终调用的是下面这个:
public HashMap(int initialCapacity, float loadFactor) {    // 参数校验 ...    this.loadFactor = loadFactor;    this.threshold = tableSizeFor(initialCapacity);}
/** * Returns a power of two size for the given target capacity. */static final int tableSizeFor(int cap) {    int n = -1 >>> Integer.numberOfLeadingZeros(cap - 1);    return (n < 0) ? 1 : (n >= MAXIMUM_CAPACITY) ? MAXIMUM_CAPACITY : n + 1;}
代码很简单,唯一需要搞清楚的就是为什么要调用tableSizeFor方法把初始化容器大小要处理成 2 的次方?
Chatgpt 给出的答案如下:
 
 构造器还支持传入一个 Map 来实例化
写一个简单的 junit case 去 debug 一下
@Testvoid testHashMapInstance() {    HashMap<Object, Object> oldMap = new HashMap<>();    oldMap.put("name", "sourceCodeExplorer");    oldMap.put("age", 18);    oldMap.put("hobby", new String[]{"sport,reading"});    System.out.println(oldMap);
    HashMap<Object, Object> newMap = new HashMap<>(oldMap);    System.out.println(newMap);
}
public HashMap(Map<? extends K, ? extends V> m) {    this.loadFactor = DEFAULT_LOAD_FACTOR;    putMapEntries(m, false);}
/** * Implements Map.putAll and Map constructor. * * @param m the map * @param evict false when initially constructing this map, else * true (relayed to method afterNodeInsertion). */final void putMapEntries(Map<? extends K, ? extends V> m, boolean evict) {    int s = m.size();    if (s > 0) {        if (table == null) { // pre-size            float ft = ((float)s / loadFactor) + 1.0F;            int t = ((ft < (float)MAXIMUM_CAPACITY) ?                     (int)ft : MAXIMUM_CAPACITY);            if (t > threshold)                threshold = tableSizeFor(t);        }        else if (s > threshold)            resize();                // 拷贝原map的所有节点到新的map中        for (Map.Entry<? extends K, ? extends V> e : m.entrySet()) {            K key = e.getKey();            V value = e.getValue();            putVal(hash(key), key, value, false, evict);        }    }}
存值方法 put()
public V put(K key, V value) {    return putVal(hash(key), key, value, false, true);}
// 计算哈希值static final int hash(Object key) {    int h;    // key的哈希值与右移16位的结果进行异或运算来实现    // 好处:1)减少冲突,降低碰撞 2) 分布更均匀    return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);}
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)        n = (tab = resize()).length;    if ((p = tab[i = (n - 1) & hash]) == null)        tab[i] = newNode(hash, key, value, null);    else {        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)            e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);        else {            for (int binCount = 0; ; ++binCount) {                if ((e = p.next) == null) {                    p.next = newNode(hash, key, value, null);                    if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st                        treeifyBin(tab, hash);                    break;                }                if (e.hash == hash &&                    ((k = e.key) == key || (key != null && key.equals(k))))                    break;                p = e;            }        }        if (e != null) { // existing mapping for key            V oldValue = e.value;            if (!onlyIfAbsent || oldValue == null)                e.value = value;            afterNodeAccess(e);            return oldValue;        }    }    ++modCount;    if (++size > threshold)        resize();    afterNodeInsertion(evict);    return null;}
扩容方法 resize()
final Node<K,V>[] resize() {        Node<K,V>[] oldTab = table;        int oldCap = (oldTab == null) ? 0 : oldTab.length;        int oldThr = threshold;        int newCap, newThr = 0;        if (oldCap > 0) {            if (oldCap >= MAXIMUM_CAPACITY) {                threshold = Integer.MAX_VALUE;                return oldTab;            }            else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY &&                     oldCap >= DEFAULT_INITIAL_CAPACITY)                newThr = oldThr << 1; // double threshold        }        else if (oldThr > 0) // initial capacity was placed in threshold            newCap = oldThr;        else {               // zero initial threshold signifies using defaults            newCap = DEFAULT_INITIAL_CAPACITY;            newThr = (int)(DEFAULT_LOAD_FACTOR * DEFAULT_INITIAL_CAPACITY);        }        if (newThr == 0) {            float ft = (float)newCap * loadFactor;            newThr = (newCap < MAXIMUM_CAPACITY && ft < (float)MAXIMUM_CAPACITY ?                      (int)ft : Integer.MAX_VALUE);        }        threshold = newThr;        @SuppressWarnings({"rawtypes","unchecked"})        Node<K,V>[] newTab = (Node<K,V>[])new Node[newCap];        table = newTab;        if (oldTab != null) {            for (int j = 0; j < oldCap; ++j) {                Node<K,V> e;                if ((e = oldTab[j]) != null) {                    oldTab[j] = null;                    if (e.next == null)                        newTab[e.hash & (newCap - 1)] = e;                    else if (e instanceof TreeNode)                        ((TreeNode<K,V>)e).split(this, newTab, j, oldCap);                    else { // preserve order                        Node<K,V> loHead = null, loTail = null;                        Node<K,V> hiHead = null, hiTail = null;                        Node<K,V> next;                        do {                            next = e.next;                            if ((e.hash & oldCap) == 0) {                                if (loTail == null)                                    loHead = e;                                else                                    loTail.next = e;                                loTail = e;                            }                            else {                                if (hiTail == null)                                    hiHead = e;                                else                                    hiTail.next = e;                                hiTail = e;                            }                        } while ((e = next) != null);                        if (loTail != null) {                            loTail.next = null;                            newTab[j] = loHead;                        }                        if (hiTail != null) {                            hiTail.next = null;                            newTab[j + oldCap] = hiHead;                        }                    }                }            }        }        return newTab;    }
取值方法 get()
public V get(Object key) {    Node<K,V> e;    return (e = getNode(hash(key), key)) == null ? null : e.value;}
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) {        if (first.hash == hash && // always check first node            ((k = first.key) == key || (key != null && key.equals(k))))            return first;        if ((e = first.next) != null) {            if (first instanceof TreeNode)                return ((TreeNode<K,V>)first).getTreeNode(hash, key);            do {                if (e.hash == hash &&                    ((k = e.key) == key || (key != null && key.equals(k))))                    return e;            } while ((e = e.next) != null);        }    }    return null;}
经典错误
初始容量计算错误
需求是存 2 个元素
// 错误Map<String,Object> map =  new HashMap<>(2);map.put("name", "code");map.put("age", 28);
// 正确Map<String, Object> map = new HashMap<>((int) (2 / 0.75f + 1));map.put("name", "code");map.put("age", 28);
使用完 map 不做 clear
See Also
- 小刘源码讲师的 HashMap 视频教程:https://www.bilibili.com/video/BV1LJ411W7dP/?vd_source=3afcc36db719cf17067a572101ab4393 
版权声明: 本文为 InfoQ 作者【前行】的原创文章。
原文链接:【http://xie.infoq.cn/article/0b47c0cd97b58280bc7497165】。文章转载请联系作者。

前行
还未添加个人签名 2018-09-06 加入
还未添加个人简介










 
    
评论