在下面的代码上下文中，SemaphoreSlim的用法是否正确？它会限制最大运行线程吗？

Question

如果我将maxConcurrency作为10传递，下面的代码将最大并行任务限制为10？如何验证一次运行的任务数量？

public BlockingCollection<Task> _workTaskQueue; 

public void DequeueTask(int maxConcurrency) 
{ 
    var tasks = new List<Task>(); 
    using (SemaphoreSlim concurrencySemaphore = new SemaphoreSlim(maxConcurrency)) 
    { 
     foreach (var task in _workTaskQueue.GetConsumingEnumerable()) 
     { 
      concurrencySemaphore.Wait(); 
      if (!(task.IsCanceled) && task.Status == TaskStatus.Created) 
      { 
       task.ContinueWith((t) => { concurrencySemaphore.Release(); }); 
       tasks.Add(task); 
       task.Start(); 
      } 
     } 
    } 
    Task.WaitAll(tasks.ToArray()); 
} 

Answer 1

由于您利用my other answer改变了一点逻辑，我为您准备了测试代码。 （无需var tasks = new List<Task>();）

Random rnd = new Random(); 
int maxConcurrency = 5; 
var _workTaskQueue = new System.Collections.Concurrent.BlockingCollection<Task>(); 
for (int i = 0; i < 250; i++) 
{ 
    //Tasks running 250=500ms 
    _workTaskQueue.Add(new Task(()=> { Task.Delay(250 + rnd.Next(250)).Wait(); })); 
} 
_workTaskQueue.CompleteAdding(); 
int runningTaks = 0; 

using (SemaphoreSlim concurrencySemaphore = new SemaphoreSlim(maxConcurrency)) 
{ 
    foreach (var task in _workTaskQueue.GetConsumingEnumerable()) 
    { 
     Console.WriteLine("LoopStart: " + runningTaks); 

     await concurrencySemaphore.WaitAsync(); 

     Console.WriteLine("GotASem : " + runningTaks); 

     task.Start(); 
     Interlocked.Increment(ref runningTaks); 
     task.ContinueWith(t => 
      { 
       Interlocked.Decrement(ref runningTaks); 
       concurrencySemaphore.Release(); 
      }); 

     if (runningTaks > maxConcurrency) throw new Exception("ERROR"); 
     Console.WriteLine("LoopEnd : " + runningTaks + Environment.NewLine); 
    } 

    Console.WriteLine("Finalizing: " + runningTaks); 
    //Make sure all all tasks have ended. 
    for (int i = 0; i < maxConcurrency; i++) 
    { 
     await concurrencySemaphore.WaitAsync(); 
    } 

    Console.WriteLine("Finished: " + runningTaks); 
} 

Answer 2

否。等待呼叫（进入互斥块）缺失。