hashmap在并发情况下的问题
- 扩容时，并发读数据可能读不到
- 并发写数据可能会丢失
本文的源码展示的是1.8.0_401版本
扩容导致并发读失败
HashMap扩容的代码如下。扩容步骤
- 先申请一块更大的空间
- table变量立刻指向这个新空间
- 旧空间里的数据逐个复制到新空间（数据量大的话会比较耗时）
步骤2结束到步骤3结束的这段时间，读数据可能会失败
final Node<K,V>[] resize() {
// 1 申请新空间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 thresholdnewCap = oldThr;else { // zero initial threshold signifies using defaultsnewCap = 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;Node<K,V>[] newTab = (Node<K,V>[])new Node[newCap];// 2 table指向新空间table = newTab;// 3 旧空间的数据复制到新空间上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 orderNode<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;elseloTail.next = e;loTail = e;}else {if (hiTail == null)hiHead = e;elsehiTail.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;}
并发写数据可能会丢失
我们看put的源码, 有两个地方，如果有多个线程同时执行，则只有一个线程的put的值得到保留。
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)// 多个线程同时运行到这里的时候，只有一个线程的值保存成功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) {// 多个线程同时运行到这里的时候，只有一个线程的值保存成功p.next = newNode(hash, key, value, null);if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1sttreeifyBin(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 keyV oldValue = e.value;if (!onlyIfAbsent || oldValue == null)e.value = value;afterNodeAccess(e);return oldValue;}}++modCount;if (++size > threshold)resize();afterNodeInsertion(evict);return null;}
验证
写段程序验证我的推测，并发1000W次读写hashmap.
public class HashMapTest {public static volatile int missingCount = 0;public static void main(String[] args) {HashMap<Integer, Integer> map = new HashMap<>();int coreSize = Runtime.getRuntime().availableProcessors();ThreadPoolExecutor es = new ThreadPoolExecutor(coreSize, coreSize, 1, TimeUnit.MINUTES,new LinkedBlockingQueue<Runnable>(100),new ThreadPoolExecutor.CallerRunsPolicy());int max = 10000000;int testMissingKey = 10001;map.put(testMissingKey, 0);for (int i = 1; i < max; i++) {int finalI = i;es.submit(() -> {System.out.println(finalI);map.put(finalI, finalI);Integer value = map.get(testMissingKey);if (value == null) {missingCount++;}});}try {es.shutdown();es.awaitTermination(1, TimeUnit.MINUTES);} catch (Exception e) {e.printStackTrace();}System.out.println((max - map.size()) + "个数字缺失了");System.out.println(missingCount + "次并发读失败");}
}
程序输出, 只看最后一部分。1000W次读写，340次读失败。弄丢 901个值。
9999917
9999916
9999915
9999912
9999909
901个数字缺失了
340次并发读失败
因此并发场景 ，必须用ConcurrentHashMap