Java5 多线程(五)--CyclicBarrier同步的工具类

允许一系列的集合等待彼此,到达一个共同的障碍物点.表示大家彼此等待,大家集合好后才开始出发,分散活动后又在指定的地点集合碰面,这就好比整个公司的人员利用周末时间集体郊游一样,先各自从家里出发到公司集合后,在同时出发到公园游玩,在指定的地点后再同时开始就餐,...public static void main(String[] args) { ExecutorService service = Executors.newCachedThreadPool(); final CyclicBarrier cb = new CyclicBarrier(3); for (int i = 0; i < 3; i++) { Runnable runnable = new Runnable() { public void run() { try { Thread.sleep((long) (Math.random() * 10000)); System.out.println("线程" + Thread.currentThread().getName() + "即将到达集合地点1，当前已有" + (cb.getNumberWaiting() + 1) + "个已经到达，" + (cb.getNumberWaiting() == 2 ? "都到齐了，继续走啊" : "正在等候")); cb.await();//只有三个线程都到齐了才往下走 Thread.sleep((long) (Math.random() * 10000)); System.out.println("线程" + Thread.currentThread().getName() + "即将到达集合地点2，当前已有" + (cb.getNumberWaiting() + 1) + "个已经到达，" + (cb.getNumberWaiting() == 2 ? "都到齐了，继续走啊" : "正在等候")); cb.await(); Thread.sleep((long) (Math.random() * 10000)); System.out.println("线程" + Thread.currentThread().getName() + "即将到达集合地点3，当前已有" + (cb.getNumberWaiting() + 1) + "个已经到达，" + (cb.getNumberWaiting() == 2 ? "都到齐了，继续走啊" : "正在等候")); cb.await(); } catch (Exception e) { e.printStackTrace(); } } }; service.execute(runnable); } service.shutdown(); }

在Api有一个使用CyclicBarrier的例子:class Solver {
final int N;
final float[][] data;
final CyclicBarrier barrier;

class Worker implements Runnable {
int myRow;
Worker(int row) { myRow = row; }
public void run() {
while (!done()) {
processRow(myRow);

try {
barrier.await();//处理完一行在此等待,如果N个线程都完成了各自的行,都在此等待,然后将会执行new CyclicBarrier(N,Runnbale)里面Runnable的代码,也就是合并行mergeRows();
} catch (InterruptedException ex) {
return;
} catch (BrokenBarrierException ex) {
return;
}
}
}
}

public Solver(float[][] matrix) {
data = matrix;
N = matrix.length;
barrier = new CyclicBarrier(N,
new Runnable() {
public void run() {
mergeRows(...);
}
});
for (int i = 0; i < N; ++i)
new Thread(new Worker(i)).start();

waitUntilDone();
}
}
在这里每个工作者(Worker)线程处理这个矩阵一行,然后在这个障碍物处等待,知道矩阵的所有行都被处理了,当所有的行都被处理了提供的Runnable障碍物被执行并且合并这些行,如果合并者确定了一个合并方案,然后将会返回(return),并且每个worker将终止.
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