Java并发编程——并发下HashMap不安全问题(ConcurrentHashMap)

问题再现：并发下对Map进行add和遍历输出，出现java.util.ConcurrentModificationException异常

public class MapTest {
    public static void main(String[] args) {
        Map<String,String> map=new HashMap<>();
        for (int i = 0; i < 30; i++) {
            new Thread(()->{
                map.put(Thread.currentThread().getName(), UUID.randomUUID().toString().substring(0,5));
                System.out.println(map);
            },String.valueOf(i)).start();
        }
    }
}

原因：HashMap在并发下线程不安全
解决方式一：
使用Collections.synchronizedMap()
Map<String,String> map= Collections.synchronizedMap(new HashMap<>());

public class MapTest {
    public static void main(String[] args) {
        //Map<String,String> map=new HashMap<>();
        Map<String,String> map= Collections.synchronizedMap(new HashMap<>());
        for (int i = 0; i < 30; i++) {
            new Thread(()->{
                map.put(Thread.currentThread().getName(), UUID.randomUUID().toString().substring(0,5));
                System.out.println(map);
            },String.valueOf(i)).start();
        }
    }
}

说明：
查看synchronizedMap()，发现该方法返回SynchronizedMap<>(m)

解决方式二：
使用ConcurrentHashMap创建map
Map<String,String> map=new ConcurrentHashMap<>();

public class MapTest {
    public static void main(String[] args) {
        //Map<String,String> map=new HashMap<>();
        //Map<String,String> map= Collections.synchronizedMap(new HashMap<>());
        Map<String,String> map=new ConcurrentHashMap<>();
        for (int i = 0; i < 30; i++) {
            new Thread(()->{
                map.put(Thread.currentThread().getName(), UUID.randomUUID().toString().substring(0,5));
                System.out.println(map);
            },String.valueOf(i)).start();
        }
    }
}

说明：
查看ConcurrentHashMap源码发现，该创建方法最后调用putAll(m)方法

putAll方法中的for循环调用putVal方法

putVal方法源码：

final V putVal(K key, V value, boolean onlyIfAbsent) {
        if (key == null || value == null) throw new NullPointerException();
        int hash = spread(key.hashCode());
        int binCount = 0;
        for (Node<K,V>[] tab = table;;) {
            Node<K,V> f; int n, i, fh;
            if (tab == null || (n = tab.length) == 0)
                tab = initTable();
            else if ((f = tabAt(tab, i = (n - 1) & hash)) == null) {
                if (casTabAt(tab, i, null,
                             new Node<K,V>(hash, key, value, null)))
                    break;                   // no lock when adding to empty bin
            }
            else if ((fh = f.hash) == MOVED)
                tab = helpTransfer(tab, f);
            else {
                V oldVal = null;
                synchronized (f) {
                    if (tabAt(tab, i) == f) {
                        if (fh >= 0) {
                            binCount = 1;
                            for (Node<K,V> e = f;; ++binCount) {
                                K ek;
                                if (e.hash == hash &&
                                    ((ek = e.key) == key ||
                                     (ek != null && key.equals(ek)))) {
                                    oldVal = e.val;
                                    if (!onlyIfAbsent)
                                        e.val = value;
                                    break;
                                }
                                Node<K,V> pred = e;
                                if ((e = e.next) == null) {
                                    pred.next = new Node<K,V>(hash, key,
                                                              value, null);
                                    break;
                                }
                            }
                        }
                        else if (f instanceof TreeBin) {
                            Node<K,V> p;
                            binCount = 2;
                            if ((p = ((TreeBin<K,V>)f).putTreeVal(hash, key,
                                                           value)) != null) {
                                oldVal = p.val;
                                if (!onlyIfAbsent)
                                    p.val = value;
                            }
                        }
                    }
                }
                if (binCount != 0) {
                    if (binCount >= TREEIFY_THRESHOLD)
                        treeifyBin(tab, i);
                    if (oldVal != null)
                        return oldVal;
                    break;
                }
            }
        }
        addCount(1L, binCount);
        return null;
    }

我们发现这个putVal方法将Node<K,V>这个对象放在代码块中用synchronized修饰！