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笔记:java并发编程实践1
Java 5.0 adds ConcurrentHashMap, a replacement for synchronized hash-based Map implementations, and CopyOnWriteArrayList, a replacement for synchronized List implementations for cases where traversal is the dominant operation. The new ConcurrentMap interface adds support for common compound actions such as put-if-absent, replace, and conditional remove.
Java 5.0 also adds two new collection types, Queue and BlockingQueue. A Queue is intended to hold a set of elements temporarily while they await processing. Several implementations are provided, including ConcurrentLinkedQueue, a traditional FIFO queue, and PriorityQueue, a (non concurrent) priority ordered queue.
Java 6 adds ConcurrentSkipListMap and ConcurrentSkipListSet, which are concurrent replacements for a synchronized SortedMap or SortedSet (such as treeMap or TReeSet wrapped with synchronizedMap).
首先我们说,替换同步hashMap的实现的concurrentHashMap,例如hashTable,写到这里突然想到了一个面试题:
在不使用锁的情况下,怎么样使hashMap 的size()方法变为线程安全的? (这个也就是要你明白ConcurrentHashMap的源码的话,就比较的容易的明白面试官要问的是什么内容)
另外还有一个比较重要的事情:
concurrentHashMap 提供的size(),isEmpty() 都是近似的精确的值,并不是完全的正确的值,这也符合并发编程环境的语意。
具体是怎么实现的,我们可以看到源码中是这样描述的:
http://www.iteye.com/topic/1103980
http://www.iteye.com/topic/344876
说的比较的详细:hashmap 被分为了segment,然后在不同的segment上面进行加锁,操作。
ConcurrentHashMap的size操作也采用了一种比较巧的方式,来尽量避免对所有的Segment都加锁。前面我们提到了一个Segment中的有一个modCount变量,代表的是对Segment中元素的数量造成影响的操作的次数,这个值只增不减,size操作就是遍历了两次Segment,每次记录Segment的modCount值,然后将两次的modCount进行比较,如果相同,则表示期间没有发生过写入操作,就将原先遍历的结果返回,如果不相同,则把这个过程再重复做一次,如果再不相同,则就需要将所有的Segment都锁住,然后一个一个遍历了。
public int size() { // Try a few times to get accurate count. On failure due to // continuous async changes in table, resort to locking. final Segment<K,V>[] segments = this.segments; int size; boolean overflow; // true if size overflows 32 bits long sum; // sum of modCounts long last = 0L; // previous sum int retries = -1; // first iteration isn‘t retry try { for (;;) { if (retries++ == RETRIES_BEFORE_LOCK) { for (int j = 0; j < segments.length; ++j) ensureSegment(j).lock(); // force creation } sum = 0L; size = 0; overflow = false; for (int j = 0; j < segments.length; ++j) { Segment<K,V> seg = segmentAt(segments, j); if (seg != null) { sum += seg.modCount; int c = seg.count; if (c < 0 || (size += c) < 0) overflow = true; } } if (sum == last) break; last = sum; } } finally { if (retries > RETRIES_BEFORE_LOCK) { for (int j = 0; j < segments.length; ++j) segmentAt(segments, j).unlock(); } } return overflow ? Integer.MAX_VALUE : size; }
这个可以对应上面的那个面试题。
再说这个CopyOnWriteArray适合的场景是:CopyOnWriteArrayList适合使用在读操作远远大于写操作的场景里,比如缓存。发生修改时候做copy,新老版本分离,保证读的高性能,适用于以读为主的情况。
笔记:java并发编程实践1