保险起见,不能直接贴出出现问题的业务代码,因此将该问题简化成如下代码:

ConcurrentHashMap<Integer, Integer> map = new ConcurrentHashMap<>();
// map默认capacity 16,当元素个数达到(capacity - capacity >> 2) = 12个时会触发rehash
for (int i = 0; i < 11; i++) {
    map.put(i, i);
}

map.computeIfAbsent(12, (k) -> {
    // 这里会导致死循环 :(
    map.put(100, 100);
    return k;
});

// 其他操作

感兴趣的小伙伴可以在电脑上运行下,话不说多,先说下问题原因:当执行computeIfAbsent时,如果key对应的slot为空,此时会创建ReservationNode对象(hash值为RESERVED=-3)放到当前slot位置,然后调用mappingFunction.apply(key)生成value,根据value创建Node之后赋值到slow位置,此时完成computeIfAbsent流程。但是上述代码mappingFunction中又对该map进行了一次put操作,并且触发了rehash操作,在transfer中遍历slot数组时,依次判断slot对应Node是否为null、hash值是否为MOVED=-1、hash值否大于0(list结构)、Node类型是否是TreeBin(红黑树结构),唯独没有判断hash值为RESERVED=-3的情况,因此导致了死循环问题。

问题分析到这里,原因已经很清楚了,当时我们认为,这可能是jdk的“bug”,因此我们最后给出的解决方案是:

  1. 如果在rehash时出现了slot节点类型是ReservationNode,可以给个提示,比如抛异常;
  2. 理论上来说,mappingFunction中不应该再对当前map进行更新操作了,但是jdk并没有禁止不能这样用,最好说明下。

最后,另一个朋友看了computeIfAbsent的注释:

 1 /**
 2  * If the specified key is not already associated with a value,
 3  * attempts to compute its value using the given mapping function
 4  * and enters it into this map unless {@code null}.  The entire
 5  * method invocation is performed atomically, so the function is
 6  * applied at most once per key.  Some attempted update operations
 7  * on this map by other threads may be blocked while computation
 8  * is in progress, so the computation should be short and simple,
 9  * and must not attempt to update any other mappings of this map.
10  */
11 public V computeIfAbsent(K key, Function<? super K, ? extends V> mappingFunction)

我们发现,其实人家已经知道了这个问题,还特意注释说明了。。。我们还是too yong too simple啊。至此,ConcurrentHashMap死循环问题告一段落,还是要遵循编码规范,不要在mappingFunction中再对当前map进行更新操作。其实ConcurrentHashMap死循环不仅仅出现在上述讨论的场景中,以下场景也会触发,原因和上述讨论的是一样的,代码如下,感兴趣的小伙伴也可以本地跑下:

1 ConcurrentHashMap<Integer, Integer> map = new ConcurrentHashMap<>();
2 map.computeIfAbsent(12, (k) -> {
3     map.put(k, k);
4     return k;
5 });
6
7 System.out.println(map);
8 // 其他操作

最后,一起跟着computeIfAbsent源码来分下上述死循环代码的执行流程,限于篇幅,只分析下主要流程代码:

 1 public V computeIfAbsent(K key, Function<? super K, ? extends V> mappingFunction) {
 2     if (key == null || mappingFunction == null)
 3         throw new NullPointerException();
 4     int h = spread(key.hashCode());
 5     V val = null;
 6     int binCount = 0;
 7     for (Node<K,V>[] tab = table;;) {
 8         Node<K,V> f; int n, i, fh;
 9         if (tab == null || (n = tab.length) == 0)
10             tab = initTable();
11         else if ((f = tabAt(tab, i = (n - 1) & h)) == null) {
12             Node<K,V> r = new ReservationNode<K,V>();
13             synchronized (r) {
14                 // 这里使用synchronized针对局部对象意义不大,主要是下面的cas操作保证并发问题
15                 if (casTabAt(tab, i, null, r)) {
16                     binCount = 1;
17                     Node<K,V> node = null;
18                     try {
19                         // 这里的value返回可能为null呦
20                         if ((val = mappingFunction.apply(key)) != null)
21                             node = new Node<K,V>(h, key, val, null);
22                     } finally {
23                         setTabAt(tab, i, node);
24                     }
25                 }
26             }
27             if (binCount != 0)
28                 break;
29         }
30         else if ((fh = f.hash) == MOVED)
31             tab = helpTransfer(tab, f);
32         else {
33             boolean added = false;
34             synchronized (f) {
35                 // 仅仅判断了node.hash >=0和node为TreeBin类型情况,未判断`ReservationNode`类型
36                 // 扩容时判断和此处类似
37                 if (tabAt(tab, i) == f) {
38                     if (fh >= 0) {
39                         binCount = 1;
40                         for (Node<K,V> e = f;; ++binCount) {
41                             K ek; V ev;
42                             if (e.hash == h &&
43                                 ((ek = e.key) == key ||
44                                  (ek != null && key.equals(ek)))) {
45                                 val = e.val;
46                                 break;
47                             }
48                             Node<K,V> pred = e;
49                             if ((e = e.next) == null) {
50                                 if ((val = mappingFunction.apply(key)) != null) {
51                                     added = true;
52                                     pred.next = new Node<K,V>(h, key, val, null);
53                                 }
54                                 break;
55                             }
56                         }
57                     }
58                     else if (f instanceof TreeBin) {
59                         binCount = 2;
60                         TreeBin<K,V> t = (TreeBin<K,V>)f;
61                         TreeNode<K,V> r, p;
62                         if ((r = t.root) != null &&
63                             (p = r.findTreeNode(h, key, null)) != null)
64                             val = p.val;
65                         else if ((val = mappingFunction.apply(key)) != null) {
66                             added = true;
67                             t.putTreeVal(h, key, val);
68                         }
69                     }
70                 }
71             }
72             if (binCount != 0) {
73                 if (binCount >= TREEIFY_THRESHOLD)
74                     treeifyBin(tab, i);
75                 if (!added)
76                     return val;
77                 break;
78             }
79         }
80     }
81     if (val != null)
82         // 计数统计&阈值判断+扩容操作
83         addCount(1L, binCount);
84     return val;
85 }

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ConcurrentHashMap竟然也有死循环问题?-LMLPHP 

06-23 16:20