Question

我正在计算股票市场指数，这个指数发生了变化，因此我决定将其分开，然后我们决定将其分开。＃34; core＆＃34; （至少24％虚拟可用核心之一）我的服务器（运行2 * E5-2640）导致到目前为止我只有5％的CPU负载，所以我有很多可用的CPU功率。

由于我需要对股市指数变化做出快速反应，我决定使用无锁代码。这是我将在现实生活中使用的例子：

using System;
using System.Diagnostics;
using System.Threading;
using System.Threading.Tasks;

namespace TestBlockingCollection2
{
    class Program
    {

        static void Main(string[] args)
        {
            Stopwatch sw = Stopwatch.StartNew();
            int i = 0;

            Task.Factory.StartNew(() =>
            {
                while (true)
                {
                    Console.WriteLine("Adding " + (i + 1));
                    sw = Stopwatch.StartNew();
                    i++;
                    Thread.Sleep(1000);
                }
            }, TaskCreationOptions.LongRunning);

            Task.Factory.StartNew(() =>
                                      {
                int prevI = 0;
                while (true)
                {
                    if (prevI < i)
                    {
                        sw.Stop();
                        long microseconds = sw.ElapsedTicks / (Stopwatch.Frequency / (1000L * 1000L));
                        Console.WriteLine("Received " + i + ". Spent " + microseconds + " microseconds.");
                        prevI = i;
                    }
                }
            }, TaskCreationOptions.LongRunning);

            while (true)
            {
                Thread.Sleep(1000);
            }
        }
    }
}

输出：

C:\bin\testBlockingCollection2>TestBlockingCollection2.exe
Adding 1
Received 1. Spent 1 microseconds.
Adding 2
Received 2. Spent 5 microseconds.
Adding 3
Received 3. Spent 2 microseconds.
Adding 4
Received 4. Spent 2 microseconds.
Adding 5
Received 5. Spent 2 microseconds.
Adding 6
Received 6. Spent 4 microseconds.
Adding 7
Received 7. Spent 2 microseconds.
Adding 8
Received 8. Spent 2 microseconds.
Adding 9
Received 9. Spent 5 microseconds.
Adding 10
Received 10. Spent 2 microseconds.
Adding 11
Received 11. Spent 2 microseconds.
Adding 12
Received 12. Spent 2 microseconds.
Adding 13
Received 13. Spent 2 microseconds.
Adding 14
Received 14. Spent 2 microseconds.
Adding 15
Received 15. Spent 2 microseconds.
Adding 16
Received 16. Spent 3 microseconds.
Adding 17
Received 17. Spent 5 microseconds.
Adding 18
Received 18. Spent 2 microseconds.
Adding 19
Received 19. Spent 2 microseconds.

与使用BlockingCollection what better do we have to asynchronous execute method than BlockingCollection?

的18微秒相比，平均2微秒相当惊人

我真的想要这个微秒，因为我有很多可用的CPU功率，并且在HFT微秒内很重要。

所以问题很简单 - 你看到我的例子有什么问题吗？我没有写过无锁代码，所以我想我可能会犯愚蠢的错误。我不使用自旋锁，因为我假设原始更改（int / double / decimal）是＆＃34; atomic＆＃34;所以在第二个线程中，我总是收到先前值或新值，但没有别的。

即。如果在一个线程中我将int x从100改为555，在第二个线程中我将有100或555，但从不500或155或其他任何东西。

优化将股票市场指数传递给其他线程的无锁代码

0 个答案: