分而治之的 CyclicBarrier

案例

    public static void main(String[] args) throws Exception{
        Map<String,Object> map = new A().test();
        for(Map.Entry<String, Object> e : map.entrySet()){
            System.out.println(e.getKey()+" : "+e.getValue());
        }

    }

    public Map<String,Object> test()throws Exception{
        final boolean[] flag = {false};
        Map<String, Object> map = new HashMap<>();

        CyclicBarrier barrier = new CyclicBarrier(3, new Runnable() {
            @Override
            public void run() {
                System.out.println("所有线程都执行完了，合并返回结果");
                flag[0] = true;
            }
        });

        // 异步调用三个类处理业务逻辑
        new Thread(()->{
            try {
                Thread.sleep(1000);
                System.out.println("线程一执行 自己的一部分工作...");
                map.put("code",1);
                barrier.await();
                System.out.println("线程一执行完");
            }catch (Exception e){
                e.printStackTrace();
            }
        }).start();

        new Thread(()->{
            try {
                Thread.sleep(1000);
                System.out.println("线程二执行 自己的一部分工作...");
                map.put("data",null);
                barrier.await();
                System.out.println("线程二执行完");
            }catch (Exception e){
                e.printStackTrace();
            }
        }).start();

        new Thread(()->{
            try {
                Thread.sleep(1000);
                System.out.println("线程三执行 自己的一部分工作...");
                map.put("msg", "删除成功");
                barrier.await();
                System.out.println("线程三执行完");
            }catch (Exception e){
                e.printStackTrace();
            }
        }).start();

        // 等三个任务执行完
        while(!flag[0]){
            //Thread.sleep(100);
        }
        return map;
    }

源码阅读

创建
    public CyclicBarrier(int parties, Runnable barrierAction) {
        if (parties <= 0) throw new IllegalArgumentException();
        // 就是将设置的 parties 和 runnable 存起来
        this.parties = parties;
        this.count = parties;
        this.barrierCommand = barrierAction;
    }


await()
    private int dowait(boolean timed, long nanos)
        throws InterruptedException, BrokenBarrierException,
               TimeoutException {
        //1. 首先上来就加上了 ReentrantLock 锁
        final ReentrantLock lock = this.lock;
        lock.lock();
        try {
            final Generation g = generation;

            if (g.broken)
                throw new BrokenBarrierException();

            if (Thread.interrupted()) {
                breakBarrier();
                throw new InterruptedException();
            }
            // 2. 我们初始化的时候就设置了count，现在递减
            int index = --count;
            if (index == 0) {  // tripped
                // 3. 这是第一种情况，index==0 说明是最后一个线程在执行任务了
                boolean ranAction = false;
                try {
                    // 4. 获取到我们初始化传入的命令，也就是说我们的那个 run() 不是线程来着，就是一个普通方法。
                    final Runnable command = barrierCommand;
                    if (command != null)
                        // 5. 执行 run() 方法
                        command.run();
                    ranAction = true;
                    // 6. 将 count重置成 partities数量，并通过Condition.signalAll()，唤醒在队列里排队的线程，也就是调用 await() 的线程，他们会尝试获取lock锁，因为那些线程此时被唤醒过后，都会进入Lock锁的AQS锁等待队列里去。
                    nextGeneration();
                    return 0;
                } finally {
                    // 7. 唤醒后面节点
                    if (!ranAction)
                        breakBarrier();
                }
            }

            // loop until tripped, broken, interrupted, or timed out
            for (;;) {
                try {
                    if (!timed)
                    // 3. 这里就是第二中情况了
                    // Condition.await()，底层，其实是释放了当前的lock锁，触发了把当前线程加入condition等待队列里，挂起当前线程
                        trip.await();
                    else if (nanos > 0L)
                        nanos = trip.awaitNanos(nanos);
                } catch (InterruptedException ie) {
                    if (g == generation && ! g.broken) {
                        breakBarrier();
                        throw ie;
                    } else {
                        // We're about to finish waiting even if we had not
                        // been interrupted, so this interrupt is deemed to
                        // "belong" to subsequent execution.
                        Thread.currentThread().interrupt();
                    }
                }

                if (g.broken)
                    throw new BrokenBarrierException();

                if (g != generation)
                    return index;

                if (timed && nanos <= 0L) {
                    breakBarrier();
                    throw new TimeoutException();
                }
            }
        } finally {
            lock.unlock();
        }
    }

 

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