• ConcurrentModificationException


    众所周知HashMap是非线程安全的数据结构,当一个线程修改了数据的时候,势必造成另一个线程读取数据异常(因为已经和最开始读的数据不一致了),那么jdk源码当中是如何实现的呢,见下图:

    以EntrySet为例,当我们调用hashmap.entrySet.iterator();时,

    public Set<Map.Entry<K,V>> entrySet() {
            Set<Map.Entry<K,V>> es;
            return (es = entrySet) == null ? (entrySet = new EntrySet()) : es;
    }
    

      

    将会得到EntryIterator实例

    final class EntrySet extends AbstractSet<Map.Entry<K,V>> {
            public final int size()                 { return size; }
            public final void clear()               { HashMap.this.clear(); }
         //返回迭代器
            public final Iterator<Map.Entry<K,V>> iterator() {
                return new EntryIterator();
            }
    

      

    而EntryIterator extends HashIterator

    final class EntryIterator extends HashIterator
            implements Iterator<Map.Entry<K,V>> {
            public final Map.Entry<K,V> next() { return nextNode(); }
    }

       

    那么迭代器内部expectedModCount 也就被初始化为外部map的modCount

    abstract class HashIterator {
            Node<K,V> next;        // next entry to return
            Node<K,V> current;     // current entry
            int expectedModCount;  // for fast-fail
            int index;             // current slot
    
            HashIterator() {
           //初始化
                expectedModCount = modCount;
                Node<K,V>[] t = table;
                current = next = null;
                index = 0;
                if (t != null && size > 0) { // advance to first entry
                    do {} while (index < t.length && (next = t[index++]) == null);
                }
            }
    

      

    当外部数据被修改过后,modCount随之被修改

        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;
                }
            }
           //更新modCount
            ++modCount;
    

      

    最后在调用nextNode()迭代的时候将会发生并发修改异常:

            final Node<K,V> nextNode() {
                Node<K,V>[] t;
                Node<K,V> e = next;
                if (modCount != expectedModCount)
                //抛出异常
                    throw new ConcurrentModificationException();
    

      

      

  • 相关阅读:
    算法基础_递归_求杨辉三角第m行第n个数字
    算法基础_递归_返回字符串的反串
    算法基础_递归_求两个字符串的公共最长子序列长度
    算法基础_递归_全排列问题
    玩转数据结构:第7章 集合和映射
    Java8 新特性_Stream API
    一、VIP课程:互联网工程专题 04-Maven私服使用与插件开发
    Java8 新特性_Lambda 表达式
    一、VIP课程:互联网工程专题 03-Maven基本概念与核心配置
    玩转数据结构:第6章 二分搜索树
  • 原文地址:https://www.cnblogs.com/lnas01/p/10361499.html
Copyright © 2020-2023  润新知