1

第一个进入的线程如何向其他并发线程发出相同方法结束的信号?

我有一个名为 PollDPRAM() 的方法。它必须通过网络访问一些慢速硬件并刷新对象私有数据。如果相同的方法被其他线程同时调用,它们一定不能执行此行程,而是等待第一个到来的线程完成工作并简单地退出,因为数据是新鲜的(比如 10-30 毫秒前没有区别) . 在不先进入第二、第三等线程的方法中很容易检测到。我使用联锁计数器来检测并发性。

问题:我通过观察计数器 (Interlocked.Read) 来检测第一个线程的退出是一个糟糕的选择,以便在计数器减少到小于在 n>1 线程入口处检测到的值之后进行观察。选择很糟糕,因为第一个线程在离开后几乎可以立即重新进入该方法。所以 n>1 线程永远不会检测到计数器的下降。

所以问题:如何正确检测第一个进入的线程已经退出方法,即使这个第一个线程可以立即再次进入它?

谢谢

PS 一段代码

        private void pollMotorsData()
    {
        // Execute single poll with "foreground" handshaking 
        DateTime start = DateTime.Now;
        byte retryCount = 0;
        // Pick old data atomically to detect change
        uint motorsDataTimeStampPrev = this.MotorsDataTimeStamp;
        bool changeDetected = false;
        // The design goal of DPRAM is to ease the bottleneck
        // Here is a sensor if bottleneck is actually that tight
        long parallelThreads = Interlocked.Increment(ref this.motorsPollThreadCount);
        try
        {
            // For first thread entering the counter will be 1
            if (parallelThreads <= 1)
            {
                do
                {
                    // Handshake signal to DPRAM write process on controller side that host PC is reading
                    this.controller.deltaTauTcpClient.Pmac_SetBit(OFFSET_0x006A_BIT15_FOREGROUND_READ, 15, true);
                    try
                    {
                        bool canReadMotors = false;
                        byte[] canReadFrozenDataFlag = new byte[2];
                        do
                        {
                            this.controller.deltaTauTcpClient.Pmac_GetMem(OFFSET_0x006E_BIT15_FOREGROUND_DONE, canReadFrozenDataFlag);
                            canReadMotors = (canReadFrozenDataFlag[1] & 0x80) == 0x80;
                            if (canReadMotors) break;
                            retryCount++;
                            Thread.Sleep(1);
                        } while (retryCount < 10);
                        if (!canReadMotors)
                        {
                            throw new DeltaTauControllerException(this.controller, "Timeout waiting on DPRAM Foreground Handshaking Bit");
                        }
                        // The lock is meaningless in contructor as it is certainly single threaded
                        // but for practice sake the access to data should always be serialized
                        lock (motorsDataLock)
                        {
                            // Obtain fresh content of DPRAM
                            this.controller.deltaTauTcpClient.Pmac_GetMem(OFFSET_0x006A_394BYTES_8MOTORS_DATA, this.motorsData);
                            this.motorsDataBorn = DateTime.Now;
                        }
                    }
                    finally
                    {
                        // Handshake signal to DPRAM write process on controller side that host PC has finished reading
                        this.controller.deltaTauTcpClient.Pmac_SetBit(OFFSET_0x006A_BIT15_FOREGROUND_READ, 15, false);
                    }
                    // Check live change in a separate atom
                    changeDetected = this.MotorsDataTimeStamp != motorsDataTimeStampPrev;
                } while ((!changeDetected) && ((DateTime.Now - start).TotalMilliseconds < 255));
                // Assert that result is live
                if (!changeDetected)
                {
                    throw new DeltaTauControllerException(this.controller, "DPRAM Background Data timestamp is not updated. DPRAM forground handshaking failed.");
                }
            }
            else
            {
                // OK. Bottleneck ! The concurrent polls have collided 
                // Give the controller a breathe by waiting for other thread do the job
                // Avoid aggressive polling of stale data, which is not able to be written, locked by reader
                // Just wait for other thread do whole polling job and return with no action
                // because the data is milliseconds fresh
                do
                {
                    // Amount of parallel threads must eventually decrease
                    // But no thread will leave and decrease the counter until job is done
                    if (Interlocked.Read(ref this.motorsPollThreadCount) < parallelThreads)
                    {
                        // Return is possible because decreased value of concurrentThreads means that
                        // this very time other thread has finished the poll 1 millisecond ago at most
                        return;
                    }
                    Thread.Sleep(1);
                    retryCount++;
                } while ((DateTime.Now - start).TotalMilliseconds < 255);
                throw new DeltaTauControllerException(this.controller, "Timeout 255ms waiting on concurrent thread to complete DPRAM polling");
            }
        }
        finally
        {
            // Signal to other threads that work is done
            Interlocked.Decrement(ref this.motorsPollThreadCount);
            // Trace the timing and bottleneck situations
            TimeSpan duration = DateTime.Now - start;
            if (duration.TotalMilliseconds > 50 || parallelThreads > 1 || retryCount > 0)
            {
                Trace.WriteLine(string.Format("Controller {0}, DPRAM poll {1:0} ms, threads {2}, retries {3}",
                    this.controller.number,
                    duration.TotalMilliseconds,
                    parallelThreads,
                    retryCount));
            }
        }
    }
4

3 回答 3

1

同步该方法,并在该方法内部检查上次完成网络访问的时间记录,以确定是否需要再次访问。

于 2011-01-14T22:29:45.427 回答
1

您可以使用“lock”关键字支持的 C# 监视器类。

基本上你的方法可以被包裹在 lock(lockobj) { CallMethod() }

假设所有线程都在同一个进程中,这将为您提供保护。

如果您需要跨进程锁定,则需要使用互斥锁。

至于您的程序,我会考虑将静态时间戳和缓存值放入您的方法中。所以当方法进入的时候,如果timestamp在我可以接受的范围内,就返回缓存的值,否则直接执行fetch。结合锁定机制,这应该可以满足您的需要。

当然,这是假设在 C# 监视器上占用和阻塞的时间不会影响您的应用程序的性能。

更新:我已经更新了您的代码,以向您展示我对使用缓存和时间戳的含义。我假设您的“motorsData”变量是从电机轮询返回的东西,因此我没有它的变量。但是,如果我误解了,只需添加一个变量来存储从代码返回的数据。注意我没有为你做任何错误检查,所以你需要处理你的异常。

    static DateTime lastMotorPoll;
    const TimeSpan CACHE_PERIOD = new TimeSpan(0, 0, 0, 0, 250);
    private object cachedCheckMotorsDataLock = new object();

    private void CachedCheckMotorsData()
    {
        lock (cachedCheckMotorsDataLock)  //Could refactor this to perform a try enter which returns quickly if required
        {
            //If the last time the data was polled is older than the cache period, poll
            if (lastMotorPoll.Add(CACHE_PERIOD) < DateTime.Now)
            {
                pollMotorsData();
                lastMotorPoll = DateTime.Now;
            }
            else //Data is fresh so don't poll
            {
                return;
            }
        }       
    }

    private void pollMotorsData()
    {
        // Execute single poll with "foreground" handshaking 
        DateTime start = DateTime.Now;
        byte retryCount = 0;
        // Pick old data atomically to detect change
        uint motorsDataTimeStampPrev = this.MotorsDataTimeStamp;
        bool changeDetected = false;
        try
        {
            do
            {
                // Handshake signal to DPRAM write process on controller side that host PC is reading
                this.controller.deltaTauTcpClient.Pmac_SetBit(OFFSET_0x006A_BIT15_FOREGROUND_READ, 15, true);
                try
                {
                    bool canReadMotors = false;
                    byte[] canReadFrozenDataFlag = new byte[2];
                    do
                    {
                        this.controller.deltaTauTcpClient.Pmac_GetMem(OFFSET_0x006E_BIT15_FOREGROUND_DONE, canReadFrozenDataFlag);
                        canReadMotors = (canReadFrozenDataFlag[1] & 0x80) == 0x80;
                        if (canReadMotors) break;
                        retryCount++;
                        Thread.Sleep(1);
                    } while (retryCount < 10);
                    if (!canReadMotors)
                    {
                        throw new DeltaTauControllerException(this.controller, "Timeout waiting on DPRAM Foreground Handshaking Bit");
                    }
                    // Obtain fresh content of DPRAM
                    this.controller.deltaTauTcpClient.Pmac_GetMem(OFFSET_0x006A_394BYTES_8MOTORS_DATA, this.motorsData);
                    this.motorsDataBorn = DateTime.Now;
                }
                finally
                {
                    // Handshake signal to DPRAM write process on controller side that host PC has finished reading
                    this.controller.deltaTauTcpClient.Pmac_SetBit(OFFSET_0x006A_BIT15_FOREGROUND_READ, 15, false);
                }
                // Check live change in a separate atom
                changeDetected = this.MotorsDataTimeStamp != motorsDataTimeStampPrev;
            } while ((!changeDetected) && ((DateTime.Now - start).TotalMilliseconds < 255));

            // Assert that result is live
            if (!changeDetected)
            {
                throw new DeltaTauControllerException(this.controller, "DPRAM Background Data timestamp is not updated. DPRAM forground handshaking failed.");
            }
        }
    }
于 2011-01-14T22:33:30.263 回答
0

有很多不同的方法可以做到这一点。正如有人已经提到的那样,您可以使用关键部分,但是如果另一个线程阻塞,这不会给您“退出”的行为。为此,您需要某种标志。您可以使用 volatile bool 并锁定对该 bool 的访问,或者您可以使用具有单个计数的信号量。最后,您可以使用互斥锁。使用同步对象的好处是您可以执行 WaitForSingleObject 并将超时设置为 0。然后您可以检查等待是否成功(如果是,则第一个线程已退出)(在这种情况下,第一个线程是仍在运行)。

于 2011-01-14T22:50:56.783 回答