Question

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你好！我有一点问题。我只学习多线程，但还不了解一切。

我有3个线程：1和2生成一个矩阵随机，并且在每一步之后它们通过管道矩阵元素发送类似参数。 3-d线程比较它们和写入结果，之后第1步和第2步进行下一步，它再次重复5次。

我的问题：我需要同步所有踏板才能正常工作。这个问题与生产者/消费者问题类似，但这里有2个生产者和1个消费者。

public class Producer1 extends Thread{
    CyclicBarrier cbar;
    public Producer1(CyclicBarrier c){
        cbar=c;
        new Thread(this).start();
    }
    private Random generator = new Random();
    int []matrix1 = new int[1000];


    private PipedWriter out = new PipedWriter();

    public PipedWriter getPipedWriter() {
        return out;
     }

    public void run() {
        for(int i =0;i<5;i++){
            matrix1[i]= generator.nextInt(10)+10;
             System.out.println("matrix1["+i+"]= "+matrix1[i]);
             try {

                  out.write(matrix1[i]);
                  cbar.await();
                  sleep(500);
                } catch (Exception e) {
                  throw new RuntimeException(e);
                }
        }
    }
}

public class Producer2 extends Thread{
     Random generator = new Random();
     int []matrix2 = new int[1000];
     CyclicBarrier cbar;
        public Producer2(CyclicBarrier c){
            cbar=c;
            new Thread(this).start();
        }
     private PipedWriter out = new PipedWriter();

        public PipedWriter getPipedWriter() {
            return out;
         }
        public void run() {
            for(int i =0;i<5;i++){
                matrix2[i]= generator.nextInt(20)-10;
                 System.out.println("matrix2["+i+"]= "+matrix2[i]);
                 try {
                      out.write(matrix2[i]);
                      cbar.await();
                      sleep(500);
                    } catch (Exception e) {
                      throw new RuntimeException(e);
                    }
            }   
        }
}

    public class Main {
    public static void main(String[] args) throws IOException { 
     Producer1 prod = new Producer1(new CyclicBarrier(2, new Consummer(prod,prod2)));   
        // here is a problem  "prod2 cannot be resolved to a variable" 
                                                     // How can i do it work??
     Producer2 prod2 = new Producer2(new CyclicBarrier(2, new Consummer(prod,prod2)));
         CyclicBarrier cb1 = new CyclicBarrier(2, new Consummer(prod,prod2)); 

        prod.start();
        prod2.start();
    }

}

public class Consummer extends Thread{
     private PipedReader el1;
     private PipedReader el2;


      public Consummer(Producer1 sender, Producer2 sender2) throws IOException {
        el1 = new PipedReader(sender.getPipedWriter());
        el2 = new PipedReader(sender2.getPipedWriter());
      }

    public void run() {
         try {

               while (true) {

                System.out.println("Element1 : " +  el1.read()+" Element2 : " +  el2.read());
               }
            } catch (IOException e) {
              throw new RuntimeException(e);
            }

    }

}


            for example : 
i=0
1-> 10
2-> 5
3-> first is bigger =10
i++;

i=1;
1-> 3
2-> 5
3-> Second is bigger =5
i++;

i=2;
1-> 4
2-> 4
3-> both are equal = 4
i++;
.....

Answer 1

可能还简化结构的一个选项是使用BlockingQueue：

在main中，创建两个队列，每个队列传递给一个生产者，两个消费者
制作人写入他们的队列
消费者从两个队列中读取并执行其操作：

while (true) {
  int[] matrix1 = queue1.remove();
  int[] matrix2 = queue2.remove();
  // process the two 
}

Answer 2

将barrier传递给消费者并让它在打印之前等待：

           while (true) {
            barrier.wait();
            System.out.println("Element1 : " +  el1.read()+" Element2 : " +  el2.read());
           }

使用Pipe Java多线程进行通信

}

2 个答案: