Java线程操作，并行编程

Question

我正在参加有关 Java 编程的在线类(class)并遇到了这个问题，我正在研究使用线程的并行编程。

我要制作一个“四核”版本的程序，通过将总和的范围分成四等份来计算 pi，并使用四个线程。

我试图将它分成 4 个不同的线程并启动和加入每个线程。

public class quadCorePi extends Thread {

public static void main(String[] args) throws Exception {

          long startTime = System.currentTimeMillis();

          quadCorePi thread1 = new quadCorePi();
          thread1.begin = 0 ;
          thread1.end = numSteps / 4 ;

          quadCorePi thread2 = new quadCorePi();
          thread2.begin = 1 ;
          thread2.end = numSteps / 4 ;

          quadCorePi thread3 = new quadCorePi();
          thread3.begin = 2 ;
          thread3.end = numSteps / 4 ;

          quadCorePi thread4 = new quadCorePi();
          thread4.begin = numSteps / 4 ;
          thread4.end = numSteps ;



          thread1.start();
          thread2.start();
          thread3.start();
          thread4.start();

          thread1.join();
          thread2.join();
          thread3.join();
          thread4.join();

          long endTime = System.currentTimeMillis();

          double pi = step * (thread1.sum + thread2.sum + thread3.sum + thread4.sum);


          System.out.println("Value of pi: " + pi);

          System.out.println("Calculated in " +
                             (endTime - startTime) + " milliseconds");
      }

但它给了我错误的 pi 值，解释对于如何为线程拆分工作非常有帮助。

这是给出的示例代码:

public class ParallelPi extends Thread {

  public static void main(String[] args) throws Exception {

      long startTime = System.currentTimeMillis();

      ParallelPi thread1 = new ParallelPi();
      thread1.begin = 0 ;
      thread1.end = numSteps / 2 ;

      ParallelPi thread2 = new ParallelPi();
      thread2.begin = numSteps / 2 ;
      thread2.end = numSteps ;

      thread1.start();
      thread2.start();

      thread1.join();
      thread2.join();

      long endTime = System.currentTimeMillis();

      double pi = step * (thread1.sum + thread2.sum) ;

      System.out.println("Value of pi: " + pi);

      System.out.println("Calculated in " +
                         (endTime - startTime) + " milliseconds");
  }

  static int numSteps = 10000000;

  static double step = 1.0 / (double) numSteps;

  double sum ;  
  int begin, end ;

  public void run() {

      sum = 0.0 ;

      for(int i = begin ; i < end ; i++){
          double x = (i + 0.5) * step ;
          sum += 4.0 / (1.0 + x * x);
      }
  }
  }

Answer 1

您的限制不正确。这是更正后的代码:

public class QuadCorePi extends Thread {

    public static void main(String[] args) throws Exception {

        long startTime = System.currentTimeMillis();

        QuadCorePi thread1 = new QuadCorePi();
        thread1.begin = 0 ;
        thread1.end = numSteps / 4 ;

        QuadCorePi thread2 = new QuadCorePi();
        thread2.begin = numSteps / 4 ;
        thread2.end =  numSteps  / 2 ;

        QuadCorePi thread3 = new QuadCorePi();
        thread3.begin = numSteps / 2 ;
        thread3.end = 3 * numSteps / 4 ;

        QuadCorePi thread4 = new QuadCorePi();
        thread4.begin = 3 * numSteps / 4 ;
        thread4.end = numSteps ;

        thread1.start();
        thread2.start();
        thread3.start();
        thread4.start();

        thread1.join();
        thread2.join();
        thread3.join();
        thread4.join();

        long endTime = System.currentTimeMillis();

        double pi = step * (thread1.sum + thread2.sum + thread3.sum + thread4.sum) ;

        System.out.println("Value of pi: " + pi);

        System.out.println("Calculated in " +
                (endTime - startTime) + " milliseconds");
    }

    static int numSteps = 10000000;

    static double step = 1.0 / (double) numSteps;

    double sum ;
    int begin, end ;

    public void run() {

        sum = 0.0 ;

        for(int i = begin ; i < end ; i++){
            double x = (i + 0.5) * step ;
            sum += 4.0 / (1.0 + x * x);
        }
    }
}