在 ActionBlock 中等待异步 lambda

Question

我有一个带有 ActionBlock 的类 Receiver：

public class Receiver<T> : IReceiver<T>
{

  private ActionBlock<T> _receiver;

  public Task<bool> Send(T item) 
  {
     if(_receiver!=null)
        return _receiver.SendAsync(item);

     //Do some other stuff her
  }

  public void Register (Func<T, Task> receiver)
  {
    _receiver = new ActionBlock<T> (receiver);
  }

  //...
}

ActionBlock 的 Register-Action 是一个带有 await-Statement 的异步方法：

private static async Task Writer(int num)
{
   Console.WriteLine("start " + num);
   await Task.Delay(500);
   Console.WriteLine("end " + num);
}

现在我要做的是同步等待（如果设置了条件），直到操作方法完成以获得独占行为：

var receiver = new Receiver<int>();
receiver.Register((Func<int, Task) Writer);
receiver.Send(5).Wait(); //does not wait the action-await here!

问题在于“等待 Task.Delay(500);”语句被执行，“receiver.Post(5).Wait();”不再等待。

我尝试了几种变体（TaskCompletionSource、ContinueWith、...），但都不起作用。

有人知道如何解决这个问题吗？

Answer 1

ActionBlock 默认情况下将强制执行排他行为（一次只处理一项）。如果您所说的“排他行为”是其他意思，您可以在操作完成时使用TaskCompletionSource 通知您的发件人：

... use ActionBlock<Tuple<int, TaskCompletionSource<object>>> and Receiver<Tuple<int, TaskCompletionSource<object>>>
var receiver = new Receiver<Tuple<int, TaskCompletionSource<object>>>();
receiver.Register((Func<Tuple<int, TaskCompletionSource<object>>, Task) Writer);
var tcs = new TaskCompletionSource<object>();
receiver.Send(Tuple.Create(5, tcs));
tcs.Task.Wait(); // if you must

private static async Task Writer(int num, TaskCompletionSource<object> tcs)
{
  Console.WriteLine("start " + num);
  await Task.Delay(500);
  Console.WriteLine("end " + num);
  tcs.SetResult(null);
}

或者，您可以使用AsyncLock (included in my AsyncEx library)：

private static AsyncLock mutex = new AsyncLock();

private static async Task Writer(int num)
{
  using (await mutex.LockAsync())
  {
    Console.WriteLine("start " + num);
    await Task.Delay(500);
    Console.WriteLine("end " + num);
  }
}