单一线程执行

Question

的.Net标准库中我只是用手写下这段代码：

class SingleWorkerThread 
{ 
    private static readonly ILog _logger = LogManager.GetLogger(typeof(SingleWorkerThread)); 
    public readonly BlockingCollection<Action> tasks = new BlockingCollection<Action>(); 
    private readonly Thread thread; 

    private void executeThread() 
    { 
     _logger.InfoFormat("Starting execution"); 
     Action task; 
     while((task = tasks.Take()) != null) 
     { 
      _logger.InfoFormat("About to execute a task...", task); 
      task(); 
      _logger.InfoFormat("Completed executing task"); 
     } 
     _logger.InfoFormat("Finished execution"); 
    } 

    public SingleWorkerThread() 
    { 
     thread = new Thread(executeThread); 
    } 

    public void Start() 
    { 
     thread.Start(); 
    } 

    public void Terminate() 
    { 
     tasks.Add(null); 
    } 

    public void AddTask(Action a) 
    { 
     tasks.Add(a); 
    } 

    public void Join() 
    { 
     // Wait up to 2 seconds for thread to terminate. 
     thread.Join(2000); 
    } 
} 

我使用它，以确保任务的某一类的所有执行总是由一个单独的线程来完成。

我想避免“重新发明轮子”，如果可能的话 - 是否有一个标准的.Net“队列消耗线程”模块，我可以用它来代替上述模块？

如果没有，在我刚刚发布的代码中是否有任何“陷阱”（即错误）？

Answer 1

我会为此使用Microsoft的Reactive Framework（NuGet“Rx-Main”）。

这是基本的代码：

class SingleWorkerThread : IDisposable 
{ 
    private Subject<Action> _subject; 
    private IScheduler _scheduler; 
    private IDisposable _subscription; 

    private static readonly ILog _logger = LogManager.GetLogger(typeof(SingleWorkerThread)); 

    public SingleWorkerThread() 
    { 
     _scheduler = new EventLoopScheduler() 
     _subject = new Subject<Action>(); 
     _subscription = 
      _subject 
       .ObserveOn(_scheduler) 
       .Subscribe(task => 
       { 
        _logger.InfoFormat("About to execute a task...", task); 
        task(); 
        _logger.InfoFormat("Completed executing task"); 
       }); 
    } 

    public void Dispose() 
    { 
     _subscription.Dispose(); 
    } 

    public void AddTask(Action Action) 
    { 
     _subject.OnNext(Action); 
    } 
}