等待在ActionBlock

Question

異步拉姆達我有一個類接收器與ActionBlock：

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); 
    } 

    //... 
} 

記數器 - 行動的ActionBlock是一個異步方法用的await-聲明：

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

現在我想要做的是等待同步（如果條件設置），直到動作方法完成以獲得獨佔行爲：

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

問題是當「等待Task.Delay（500）;」語句被執行，「receiver.Post（5）.Wait（）;」不再等待。

我嘗試了幾種變體（TaskCompletionSource，ContinueWith，...），但它不起作用。

有沒有人有一個想法如何解決這個問題？

Answer 1

ActionBlock默認情況下會強制執行排他行爲（一次只處理一個項目）。如果你的意思是「獨佔行爲」別的東西，你可以用TaskCompletionSource通知您發送者的動作完成時：

... 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，你可以使用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); 
    } 
}