只进入第一个Thread,无法进入以下线程

Question

我想使用多个线程处理数据,但目前只有第一个线程运行,然后它完全停止.

我正在尝试打印出Threads的名称,但我只是输入了第一个Thread.即使程序继续运行,它也会停止.它去了synchpoint.await()当时的停止.

public class Solver {
  final int N;
  final float[][] data;
  final CyclicBarrier barrier;

class Worker implements Runnable {
    public int myRow;
    String name;
    public CyclicBarrier synchPoint;

    Worker(CyclicBarrier barrier, int row, String name) {
        myRow = row;
        this.name = name;
        this.synchPoint = barrier;
        this.run();
    }

    public void run() {
            System.out.println("Name: " + name);
            processRow(myRow);
            mergeRows();

            try {
                System.out.print("In SynchPoint");
                synchPoint.await();
            } catch (InterruptedException ex) {
                return;
            } catch (BrokenBarrierException ex) {
                return;
            }
    }
}

public void processRow(int numRow){
    System.out.println("In Process Row");
}

public void mergeRows(){
    System.out.println("In merge Row");
}

public Solver(float[][] matrix) {
    data = matrix;
    N = matrix.length;
    System.out.println("N: " + N);
    Runnable barrierAction =
            new Runnable() { public void run() { mergeRows(); }};
    barrier = new CyclicBarrier(N, barrierAction);

    List<Thread> threads = new ArrayList<Thread>(N);
    for (int i = 0; i < N; i++) {
        String myName = "Worker-" + i;
        Thread thread = new Thread(new Worker(barrier, i, myName));
        threads.add(thread);
        thread.start();
        System.out.println("In loop, i is: " + i);
    }

    // wait until done
    for (Thread thread : threads)
        try {
            thread.join();
        } catch (InterruptedException e) {
            e.printStackTrace();
        }
  }

  public static void doWork(){
      float[][] matrix2 = new float[6][2000];
      Solver solve = new Solver(matrix2);
  }

  public static void main(String[] args) {
      Solver.doWork();
  }
}

输出:

N:5
名称:
Process -0 在Process Row
Row中:0
数据长度:5
在合并行中
合并行在
SynchPoint中

Answer 1

你有一个讨厌的习惯,让你的构造函数除了初始化一个实例之外还能工作.Solver构造函数执行此操作并不是最终有害的,但Solver.Worker调用实例run()方法的事实是您的问题的根源.

您Worker在应用程序的主线程中运行构造函数.该构造函数调用run(),它在到达时挂起该线程synchPoint.await().当足够的其他线程到达该点时,线程将恢复,但是没有一个线程将会恢复,因为所做的一个线程是创建并启动其他线程的线程,并且它无法继续.

删除this.run()from Worker的构造函数,你应该得到更多.