第一个进入的线程如何向同一方法结束的其他并发线程发信号？
我有一个名为PollDPRAM（）的方法。它必须通过网络到一些慢速硬件和刷新对象私有数据。如果其他线程同时调用相同的方法，则它们不能执行该行程，而是等待第一个到来的线程完成作业并退出，因为数据是新鲜的（比如说10-30毫秒之前没有区别） 。 它很容易在方法中检测到第二个，第三个等线程没有首先输入。我使用Interlocked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，在计数器减少到比在n> 1个线程的入口处检测到的值减少之后观察。选择很糟糕，因为第一个线程可以在离开后几乎立即重新进入该方法。因此，n> 1个线程永远不会检测到计数器中的下降。
所以问题： 如何正确检测第一个进入的线程已经退出该方法，即使该第一个线程可以立即再次输入它？
谢谢
P.S。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));
}
}
}
答案 0 :(得分：1)
同步方法并在方法内部检查上次进行网络访问的时间记录，以确定是否需要再次完成。
答案 1 :(得分：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用lock（lockobj）{CallMethod（）}
包装假设所有线程都在同一个进程中，这将为您提供保护。
如果您需要锁定进程，则需要使用互斥锁。
至于你的程序，我会考虑将静态时间戳和缓存值放入你的方法中。因此，当方法进入时，如果时间戳在我可接受的范围内，则返回缓存的值，否则只需执行获取。结合锁定机制，这应该做你需要的。
当然，这假设在C＃监视器上占用和阻止的时间不会影响应用程序的性能。
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意义。我假设你的“motorsData”变量是从电机轮询返回的东西，因此我没有变量。但是如果我误解了，只需添加一个变量来存储从代码中返回的数据。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.");
}
}
}
答案 2 :(得分：0)
你有很多不同的方法可以做到这一点。你可以使用一个关键部分，正如有人已经提到的那样，但如果另一个线程阻塞，那么就不会给你“退出”的行为。为此你需要某种旗帜。您可以使用易变的bool并锁定该bool的访问权限，或者您可以使用单个计数的信号量。最后，您可以使用互斥锁。使用同步对象的好处是你可以做一个WaitForSingleObject并将超时设置为0.然后你可以检查等待是否成功（如果它是第一个线程已经退出）或者不是（在这种情况下第一个线程是还在跑步。）