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
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