JDK容器与并发—Map—IdentityHashMap
2016-04-20 17:36
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概述
基于数组,用==比较key的类HashMap,非线程安全。1)在System#identityHashCode哈希合理情况下,基础操作get、put等为固定时间;
2)影响性能的参数:期望最大键值对数(即键值对threshold),该参数在初始化时决定bucket数。如果map中的键值对数超过了该参数,bucket数则增长,进一步导致rehash;另外,容器视图迭代器遍历时间与bucket数成比例,如果对迭代性能或内存成本有较高要求,该参数不应该设置过大。
3)适用于做备份数据:保留拓扑结构象图转化、保留代理对象,都用到了对象的唯一性;
4)迭代器fail-fast。
数据结构
Object数组,逻辑上是环形的。长度为2的幂次方,table[i]为key,table[i+1]为该key关联的value,其中i为偶数。private transient Object[] table;
构造器
了解几个概念:最大键值对:table.length/2; 默认负载因子:2/3; 期望最大键值对数:最大键值对*负载因子,达到该值则rehash。// 无参构造,采用默认最大键值对数32,键值对阈值为21 public IdentityHashMap() { init(DEFAULT_CAPACITY); } // 带期望最大键值对数构造 public IdentityHashMap(int expectedMaxSize) { if (expectedMaxSize < 0) throw new IllegalArgumentException("expectedMaxSize is negative: " + expectedMaxSize); init(capacity(expectedMaxSize)); } // 带Map参数构造 public IdentityHashMap(Map<? extends K, ? extends V> m) { // Allow for a bit of growth this((int) ((1 + m.size()) * 1.1)); putAll(m); }
增删改查
容量调整策略
与HashMap大致相同:1)当IdentityHashMap中键值对数达到键值对阈值则对其table进行容量调整
,table容量翻倍;
2)table最大容量为1 << 30,即最大键值对数为MAXIMUM_CAPACITY = 1 << 29;
3)table容量达到最大后,如果有resize请求,则仅仅将键值对阈值调整为MAXIMUM_CAPACITY-1,不会产生新的table,还可以继续添加新的键值对,之后如果再有resize请求,则会抛出容量异常;
4)若table resize过程中产生新的table,需要将旧table的键值对重新在新table中确定bucket,再添加进来,也就是所说的hash table rehash。
private void resize(int newCapacity) { // assert (newCapacity & -newCapacity) == newCapacity; // power of 2 int newLength = newCapacity * 2; Object[] oldTable = table; int oldLength = oldTable.length; if (oldLength == 2*MAXIMUM_CAPACITY) { // can't expand any further if (threshold == MAXIMUM_CAPACITY-1) throw new IllegalStateException("Capacity exhausted."); threshold = MAXIMUM_CAPACITY-1; // Gigantic map! return; } if (oldLength >= newLength) return; Object[] newTable = new Object[newLength]; threshold = newLength / 3; // 键值对负载因子依然为2/3,threshold几乎翻倍 // rehash for (int j = 0; j < oldLength; j += 2) { Object key = oldTable[j]; if (key != null) { Object value = oldTable[j+1]; oldTable[j] = null; oldTable[j+1] = null; int i = hash(key, newLength); // 在newTable中获取bucketIndex while (newTable[i] != null) // 解决hash碰撞 i = nextKeyIndex(i, newLength); // 循环后移2位 newTable[i] = key; newTable[i + 1] = value; } } table = newTable; } private static int hash(Object x, int length) { // 不管x有无Override Object的hashCode方法,都会返回原始的Object的hashCode值 int h = System.identityHashCode(x); // Multiply by -127, and left-shift to use least bit as part of hash return ((h << 1) - (h << 8)) & (length - 1); } // 循环右移2位 private static int nextKeyIndex(int i, int len) { return (i + 2 < len ? i + 2 : 0); }
增、改
与HashMap相似,步骤:1)根据System.identityHashCode(key)获取hash码;
2)用hash码确定bucketIndex;
3)先遍历bucket中键值对,确定是否已有==key的,有则替换新value后返回;否则将key—value对用数组的方式添加进来。
public V put(K key, V value) { Object k = maskNull(key); Object[] tab = table; int len = tab.length; int i = hash(k, len); // 确定bucketIndex Object item; while ( (item = tab[i]) != null) { if (item == k) { // 用==判断key是否相同 V oldValue = (V) tab[i + 1]; tab[i + 1] = value; return oldValue; } i = nextKeyIndex(i, len); } modCount++; tab[i] = k; tab[i + 1] = value; if (++size >= threshold) resize(len); // len == 2 * current capacity. return null; }
删
步骤:1)根据System.identityHashCode(key)获取hash码;
2)用hash码确定bucketIndex;
3)遍历bucket中键值对,确定是否已有==key的,有则删除且对其后的键值对进行rehash,否则返回null
public V remove(Object key) { Object k = maskNull(key); Object[] tab = table; int len = tab.length; int i = hash(k, len); while (true) { Object item = tab[i]; if (item == k) { modCount++; size--; V oldValue = (V) tab[i + 1]; tab[i + 1] = null; tab[i] = null; closeDeletion(i); // 因删除产生的rehash return oldValue; } if (item == null) return null; i = nextKeyIndex(i, len); } } // 对所删除键值对后的键值对进行rehash private void closeDeletion(int d) { // Adapted from Knuth Section 6.4 Algorithm R Object[] tab = table; int len = tab.length; // Look for items to swap into newly vacated slot // starting at index immediately following deletion, // and continuing until a null slot is seen, indicating // the end of a run of possibly-colliding keys. Object item; for (int i = nextKeyIndex(d, len); (item = tab[i]) != null; i = nextKeyIndex(i, len) ) { // The following test triggers if the item at slot i (which // hashes to be at slot r) should take the spot vacated by d. // If so, we swap it in, and then continue with d now at the // newly vacated i. This process will terminate when we hit // the null slot at the end of this run. // The test is messy because we are using a circular table. int r = hash(item, len); if ((i < r && (r <= d || d <= i)) || (r <= d && d <= i)) { tab[d] = item; tab[d + 1] = tab[i + 1]; tab[i] = null; tab[i + 1] = null; d = i; } } }
查
步骤:1)根据System.identityHashCode(key)获取hash码;
2)用hash码确定bucketIndex;
3)遍历bucket中键值对,确定是否已有==key的,有则返回关联的value,否则返回null
public V get(Object key) { Object k = maskNull(key); Object[] tab = table; int len = tab.length; int i = hash(k, len); while (true) { Object item = tab[i]; if (item == k) return (V) tab[i + 1]; if (item == null) return null; i = nextKeyIndex(i, len); } }
迭代器
利用数组特性遍历,IdentityHashMapIterator为基础迭代器,其删除过程中需要对其后的键值对rehash。特性
解决hash碰撞1)IdentityHashMap的table数组的长度为2的幂次方;
2)根据System.identityHashCode(key)获取hash码,用hash码确定bucketIndex;
3)用数组解决hash碰撞问题。
其与HashMap解决hash碰撞的方式是不一样的。
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