为什么这段Java代码没有同时运行

时间:2013-09-08 08:41:52

标签: java multithreading concurrency sieve-of-eratosthenes

我写过了Eratosthenes的筛子,它应该是平行的,但事实并非如此。当我增加线程数时,计算时间不会越来越少。有什么想法吗?

主要课程

import java.util.Date;
import java.util.concurrent.ExecutorService;
import java.util.concurrent.Executors;

public class ConcurrentTest {
    public static void main(String[] args) throws InterruptedException {
        Sieve task = new Sieve();
        int x = 1000000;
        int threads = 4;
        task.setArray(x);
        Long beg = new Date().getTime();
        ExecutorService exec = Executors.newCachedThreadPool();
        for (int i = 0; i < threads; i++) {
            exec.execute(task);
        }
        exec.shutdown();
        Long time = 0L;
    // Main thread is waiting until all threads are terminated
    // ( it means that computing is done)
        while (true)
            if (exec.isTerminated()) {
                time = new Date().getTime() - beg;
                break;
            }

        System.out.println("Time is " + time);
    }
}

Sieve class

import java.util.concurrent.ConcurrentHashMap;

public class Sieve implements Runnable {
    private ConcurrentHashMap<Integer, Boolean> array = 
                       new ConcurrentHashMap<Integer, Boolean>();
    private int x;
    public void run() {
        while(true){
    // I am getting synchronized number to check if it's prime
            int n = getCounter();
    // If no more numbers to check, stop loop
            if( n == -1)
                break;
    // If HashMap contains number, we can further
            if(!array.containsKey(n))continue;
            for (int i = 2 * n; i <= x; i += n) {
    // Compound numbers are removed from HashMap, Eg. 6, 12 and much more.
                    array.remove(i);
            }
        }
    }
    private synchronized int getCounter(){
        if( counter < x)
            return counter++;
        else return -1;
    }
    public void setArray(int x) {
        this.x = x;
        for (int i = 2; i <= x; i++)
            array.put(i, false);
    }
}

我用不同数量的线程进行了一些测试。结果如下:

Nr of threads 1    Time is 1850, 1795, 1825
Nr of threads 2    Time is 1845, 1836, 1814
Nr of threads 3    Time is 1767, 1820, 1756
Nr of threads 4    Time is 1732, 1840, 2083
Nr of threads 5    Time is 1791, 1795, 1803
Nr of threads 6    Time is 1825, 1728, 1707
Nr of threads 7    Time is 1754, 1729, 1686
Nr of threads 8    Time is 1760, 1717, 1817
Nr of threads 9    Time is 1721, 1699, 1673
Nr of threads 10   Time is 1661, 1722, 1718

1 个答案:

答案 0 :(得分:4)

  

当我增加线程数时,计算时间就没有了   下

tl; dr :您的问题规模太小。如果将x增加到10000000,差异将变得更加明显。不过,它们不会是你所期待的。

我在八核机器上尝试了你的代码,稍作修改:

  1. 为了计时,我在日期使用了System.nanoTime()而不是getTime()。
  2. 我使用ExecutorService的awaitTermination方法而不是spinloop来检查运行的结束。
  3. 我尝试使用fixed thread poolcached thread poolfork join pool启动Sieve任务,并比较线程变量的不同值的结果。

    我在x = 10000000的机器上看到以下结果(以毫秒为单位):

        Thread count      = 1    2    4    8    16
        Fixed thread pool = 5451 3866 3639 3227 3120
        Cached thread pool= 5434 3763 3709 3258 3078
        Fork-join pool    = 6732 3670 3735 3190 3102
    

    这些结果向我们展示了从单个执行线程更改为两个线程的明显好处。但是,额外线程的好处会迅速下降。有一个有趣的高原,从2到4个线程,边际效益高达16个。

    此外,您还可以看到不同的线程机制具有不同的初始开销:我没想到Fork-Join池的启动成本比其他机制要多得多。

    因此,正如所写的那样,对于小而非平凡的问题集,你不应该期望通过两个线程获益。

    如果您想增加其他线程的好处,那么您需要查看当前的实现。例如,当我使用AtomicInteger从同步的getCounter()切换到incrementAndGet()时,我消除了synchronized方法的开销。结果是我的所有四个线程数都下降了1000毫秒。