【图解】面试题:ConcurrentHashMap是如何保证线程安全的

注意:JDK1.7与JDK1.8中的ConcurrentHashMap主要延续HashMap的设计与思想,是在其基础上进行的相应优化

1.JDK1.7中的底层实现原理
(1)JDK1.7ConcurrentHashMap的底层结构,延续HashMap的底层设计,采用数组+链表

(2)不同的是:ConcurrentHashMap中的数组被分为大数组和小数组,大数组是Segment,小数组是HashEntry
Segment本身是基于ReentrantLock可重入锁来实现加锁和释放锁,这样就能保证多线程同时访问ConcurrentHashMap的时候,同一时间只能有一个线程操作对应的节点,以此来保证线程安全

总结:这里基于Segment加锁和释放锁,因此称之为分段锁或分片锁

【图解】面试题:ConcurrentHashMap是如何保证线程安全的_第1张图片

2.JDK1.8中的底层实现原理
(1)JDK1.8的ConcurrentHashMap底层采用的是数组+链表+红黑树,这里对JDK1.7进行了优化,在链表长度大于8并且数组长度大于64时链表就会转化为红黑树

(2)JDK1.8中保留了Segment的定义,这仅仅是为了保证序列化时的兼容性,不再有任何结构上的用途

(3)JDK1.8中主要使用volatile+CAS或者是synchronized的方法来实现的,以此来保证线程安全
(i)源码中,添加元素时,首先会判断容器是否为空,为空,就会使用volatile加CAS来初始化
(ii)容器不为空,就会根据存储的元素计算该位置是否为空,如果计算结果位置的结果为空,就会使用CAS来设计该节点,不为空,就会使用synchronized来加锁实现,然后遍历桶中的数据并且替换或者新增节点到桶中,最后判断是否需要转换红黑树

总结:JDK1.8中ConcurrentHashMap是通过对头节点加锁来保证线程安全,降低锁粒度,发生hash冲突和加锁的频率也更低了

【图解】面试题:ConcurrentHashMap是如何保证线程安全的_第2张图片
附上JDK1.8源码

 /**
     * Maps the specified key to the specified value in this table.
     * Neither the key nor the value can be null.
     *
     * 

The value can be retrieved by calling the {@code get} method * with a key that is equal to the original key. * * @param key key with which the specified value is to be associated * @param value value to be associated with the specified key * @return the previous value associated with {@code key}, or * {@code null} if there was no mapping for {@code key} * @throws NullPointerException if the specified key or value is null */ public V put(K key, V value) { return putVal(key, value, false); } /** Implementation for put and putIfAbsent */ 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; }

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