本篇将详细分析JDK8中HashMap的源码及其工作原理
版本
本篇使用版本
1 2 3 java version "1.8.0_151" Java(TM) SE Runtime Environment (build 1.8.0_151-b12) Java HotSpot(TM) 64-Bit Server VM (build 25.151-b12, mixed mode)
HashMap继承关系
常量
1 2 3 4 5 6 7 8 9 10 11 12 static final int DEFAULT_INITIAL_CAPACITY = 1 << 4 ; static final int MAXIMUM_CAPACITY = 1 << 30 ;static final float DEFAULT_LOAD_FACTOR = 0.75f ;static final int TREEIFY_THRESHOLD = 8 ;static final int UNTREEIFY_THRESHOLD = 6 ; static final int MIN_TREEIFY_CAPACITY = 64 ;
变量
1 2 3 4 5 6 7 8 9 10 11 12 transient Node<K,V>[] table;transient Set<Map.Entry<K,V>> entrySet;transient int size;transient int modCount;int threshold;final float loadFactor;
构造方法
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 public HashMap () { this .loadFactor = DEFAULT_LOAD_FACTOR; } public HashMap (int initialCapacity) { this (initialCapacity, DEFAULT_LOAD_FACTOR); } public HashMap (int initialCapacity, float loadFactor) { if (initialCapacity < 0 ) throw new IllegalArgumentException("Illegal initial capacity: " + initialCapacity); if (initialCapacity > MAXIMUM_CAPACITY) initialCapacity = MAXIMUM_CAPACITY; if (loadFactor <= 0 || Float.isNaN(loadFactor)) throw new IllegalArgumentException("Illegal load factor: " + loadFactor); this .loadFactor = loadFactor; this .threshold = tableSizeFor(initialCapacity); } public HashMap (Map<? extends K, ? extends V> m) { this .loadFactor = DEFAULT_LOAD_FACTOR; putMapEntries(m, false ); }
方法
tableSizeFor
(int,float)构造初始化时调用的,作用是给定一个参数,返回大于输入参数且最近的2的整数次幂的数。源码如下
1 2 3 4 5 6 7 8 9 static final int tableSizeFor (int cap) { int n = cap - 1 ; n |= n >>> 1 ; n |= n >>> 2 ; n |= n >>> 4 ; n |= n >>> 8 ; n |= n >>> 16 ; return (n < 0 ) ? 1 : (n >= MAXIMUM_CAPACITY) ? MAXIMUM_CAPACITY : n + 1 ; }
详解 :
假设我们现在输入10,那么n的初始值为9,转换为二进制就是 0000…1001
右移一位变成0000…0100,再位或变成0000…1101
右移两位变成0000…0011,再位或变成0000…1111
右移四位变成0000…0000,再位或变成0000…1111 ,我们发现,到这里之后,无论如何运算,n都不会再改变了,最后n的结果为15
最后一句return表示 当n小于0返回1;大于最大容量时,返回最大容量。如果正常合法,则返回n+1
解释一下这个方法,其实就是利用位运算,将传入参数的最高位前一位变为0且后面的所有位都为1。这样当n+1时,最高位前一位变成1,其他位全为0,这样就得到了大于这个数字且最近的2的整数次幂。
备注 :
为什么要先减一int n = cap - 1
当输入参数是本就是2的幂次方时,位运算计算出来的就是大于这个参数且最近的2的幂次方。其实就是为了解决当参数是2的整数次幂时的计算错误问题。
为什么要加一,因为位运算算出来的结果必然为001…1 ,…中全是1,加一后才能得到2的整数次幂。
putMapEntries
map构造时调用的,传入一个map类型,初始化数组容量并将入参Map的KV键值对添加到当前的HashMap中。
源码如下:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 final void putMapEntries (Map<? extends K, ? extends V> m, boolean evict) { int s = m.size(); if (s > 0 ) { if (table == null ) { 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(); 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); } } }
put
1 2 3 4 public V put (K key, V value) { return putVal(hash(key), key, value, false , true ); }
hash
1 2 3 4 5 static final int hash (Object key) { int h; return (key == null ) ? 0 : (h = key.hashCode()) ^ (h >>> 16 ); }
putVal
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 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 ) treeifyBin(tab, hash); break ; } if (e.hash == hash && ((k = e.key) == key || (key != null && key.equals(k)))) break ; p = e; } } if (e != null ) { 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resize
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 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 ; } else if (oldThr > 0 ) newCap = oldThr; else { 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 { 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; }
获取方法get
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 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 && ((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 ; }
图解
存储结构
put流程图
扩容
备注
有关红黑树的知识有点多,决定单独写一篇来说红黑树和HashMap中的红黑树实现