CyclicBarrier中await(int time,TimeUnit unit)超时问题

Talk is cheap.Show the code！

public class MyCyclicBarrier_1 extends Thread{
    private CyclicBarrier cyclicBarrier;

    public MyCyclicBarrier_1(CyclicBarrier cyclicBarrier) {
        this.cyclicBarrier = cyclicBarrier;
    }

    @Override
    public void run() {
        System.out.println("Thread start."+Thread.currentThread().getName());
        try {
            TimeUnit.SECONDS.sleep(2);  //业务模拟code
            System.out.println("Thread "+Thread.currentThread().getName()+" is waiting for the other Threads."+
                    "\n\t\t\t\tIt's parties is "+cyclicBarrier.getParties()+
                    "\n\t\t\t\tWaiting for "+cyclicBarrier.getNumberWaiting()+" Threads");
            cyclicBarrier.await(3,TimeUnit.SECONDS);
        } catch (InterruptedException | BrokenBarrierException | TimeoutException e) {
            e.printStackTrace();
        }
        System.out.println("Thread end."+Thread.currentThread().getName());
    }
}

测试代码 

public class TestCyclicbarrier_1 {
    public static void main(String[] args) {
        int length = 5;
        long start = System.currentTimeMillis();
        CyclicBarrier cyclicBarrierWithRunnable = new CyclicBarrier(length,()->{
            System.out.println("the final reach Thread is "+Thread.currentThread().getName());
            long end = System.currentTimeMillis();
            System.out.println("cost totally :"+(end-start)/1000+"s");
        });
        for (int i=0;i<length;i++){
            if (i!=4){
                new MyCyclicBarrier_1(cyclicBarrierWithRunnable).start();
            }else {
                try {
                    TimeUnit.SECONDS.sleep(2);    //若是这里改成3s会抛出超时异常
                    new MyCyclicBarrier_1(cyclicBarrierWithRunnable).start();
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
            }
        }
    }
}

输出：

Thread start.Thread-1
Thread start.Thread-0
Thread start.Thread-2
Thread start.Thread-3
Thread Thread-0 is waiting for the other Threads.
                It's parties is 5
                Waiting for 0 Threads
Thread Thread-3 is waiting for the other Threads.
                It's parties is 5
                Waiting for 0 Threads
Thread start.Thread-4
Thread Thread-1 is waiting for the other Threads.
                It's parties is 5
                Waiting for 0 Threads
Thread Thread-2 is waiting for the other Threads.
                It's parties is 5
                Waiting for 1 Threads
Thread Thread-4 is waiting for the other Threads.
                It's parties is 5
                Waiting for 4 Threads
the final reach Thread is Thread-4
cost totally :4s
Thread end.Thread-4
Thread end.Thread-0
Thread end.Thread-3
Thread end.Thread-2
Thread end.Thread-1

问题一眼就能看出来了，0至3这三条线程在2s的时候就已经“到达了”屏障点，CyclicBarrier.await(3s)。为什么线程4（最后执行的这条线程）沉睡了2s后再执行2秒的业务代码（共4秒），不会使得CyclicBarrier.await(3s)抛出超时异常呢？

已分析出答案，此贴可结。