c# - 为什么 volatile 和 MemoryBarrier 不阻止操作重新排序？

Question

如果我正确理解 volatile 和 MemoryBarrier 的含义，那么下面的程序将永远无法显示任何结果。

每次我运行它时，它都会捕获写入操作的重新排序。我在 Debug 或 Release 中运行它并不重要。将它作为 32 位或 64 位应用程序运行也没有关系。

为什么会这样？

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

    namespace FlipFlop
    {
        class Program
        {
            //Declaring these variables as volatile should instruct compiler to 
            //flush all caches from registers into the memory.
            static volatile int a;
            static volatile int b;

            //Track a number of iteration that it took to detect operation reordering.
            static long iterations = 0;

            static object locker = new object();

            //Indicates that operation reordering is not found yet.
            static volatile bool continueTrying = true;

            //Indicates that Check method should continue.
            static volatile bool continueChecking = true;

            static void Main(string[] args)
            {
                //Restarting test until able to catch reordering.
                while (continueTrying)
                {
                    iterations++;
                    var checker = new Task(Check);
                    var writter = new Task(Write);
                    lock (locker)
                    {
                        continueChecking = true;
                        checker.Start();

                    }
                    writter.Start();
                    checker.Wait();
                    writter.Wait();
                }
                Console.ReadKey();
            }

            static void Write()
            {
                //Writing is locked until Main will start Check() method.
                lock (locker)
                {
                    //Using memory barrier should prevent opration reordering.
                    a = 1;
                    Thread.MemoryBarrier();
                    b = 10;
                    Thread.MemoryBarrier();
                    b = 20;
                    Thread.MemoryBarrier();
                    a = 2;

                    //Stops spinning in the Check method.
                    continueChecking = false;
                }
            }

            static void Check()
            {
                //Spins until finds operation reordering or stopped by Write method.
                while (continueChecking)
                {
                    int tempA = a;
                    int tempB = b;

                    if (tempB == 10 && tempA == 2)
                    {
                        continueTrying = false;
                        Console.WriteLine("Caught when a = {0} and b = {1}", tempA, tempB);
                        Console.WriteLine("In " + iterations + " iterations.");
                        break;
                    }
                }
            }
        }
    }

Answer 1

你没有在测试之间清理变量，所以(除了第一个)最初 a 是 2 而 b 是 20 - before Write 已完成任何。

Check 可以获取 a 的initial 值(所以 tempA 是2)，然后Write就可以进去了，尽量把b改成10。

现在 Check 读取 b(所以 tempB 是 10)。

等等。无需重新订购即可重现。

在运行之间将 a 和 b 重置为 0，我希望它会消失。

编辑:确认； “按原样”我几乎立即得到了问题(<2000 次迭代)；但通过添加:

while (continueTrying)
{
    a = b = 0; // reset <======= added this

然后它可以毫无问题地循环任意时间。

或者作为一个流程:

Write                   A=  B=        Check

(except first run)      2   20
                                      int tempA = a;
a = 1;                  1   20
Thread.MemoryBarrier();
b = 10;                 1   10
                                      int tempB = b;