是否有默认方式获得成功完成的第一个任务？

Question

让我们说我有几个任务:

void Sample(IEnumerable<int> someInts)
{
    var taskList = someInts.Select(x => DownloadSomeString(x));
}

async Task<string> DownloadSomeString(int x) {...}

我想得到第一个成功任务的结果.所以,基本的解决方案是写下这样的东西:

var taskList = someInts.Select(x => DownloadSomeString(x));
string content = string.Empty;
Task<string> firstOne = null;
while (string.IsNullOrWhiteSpace(content)){
    try
    {
        firstOne = await Task.WhenAny(taskList);
        if (firstOne.Status != TaskStatus.RanToCompletion)
        {
            taskList = taskList.Where(x => x != firstOne);
            continue;
        }
        content = await firstOne;
    }
    catch(...){taskList = taskList.Where(x => x != firstOne);}
}

但是这个解决方案似乎运行N+( - N1)+ .. + K任务.哪里N是someInts.Count和K在任务第一次成功的任务的位置,这样它重新运行所有任务,除了一个由WhenAny抓获.那么,有没有办法让第一个任务成功完成并运行最多的N任务？(如果成功的任务将是最后一个)

Answer 1

您需要做的就是创建一个TaskCompletionSource,为每个任务添加一个延续,并在第一个任务成功完成时设置它:

public static Task<T> FirstSuccessfulTask<T>(IEnumerable<Task<T>> tasks)
{
    var taskList = tasks.ToList();
    var tcs = new TaskCompletionSource<T>();
    int remainingTasks = taskList.Count;
    foreach (var task in taskList)
    {
        task.ContinueWith(t =>
            {
                if (task.Status == TaskStatus.RanToCompletion)
                    tcs.TrySetResult(t.Result);
                else
                if (Interlocked.Decrement(ref remainingTasks) == 0)
                    tcs.SetException(new AggregateException(tasks.SelectMany(t1 => t1.Exception.InnerExceptions)));
            });
    }
    return tcs.Task;
}

没有结果的任务版本:

public static Task FirstSuccessfulTask(IEnumerable<Task> tasks)
{
    var taskList = tasks.ToList();

    var tcs = new TaskCompletionSource<bool>();

    int remainingTasks = taskList.Count;

    foreach (var task in taskList)
    {
        task.ContinueWith(t =>
        {
            if (task.Status == TaskStatus.RanToCompletion)
                tcs.TrySetResult(true);
            else
                if (Interlocked.Decrement(ref remainingTasks) == 0)
                tcs.SetException(new AggregateException(
                    tasks.SelectMany(t1 => t1.Exception.InnerExceptions)));
        });
    }

    return tcs.Task;
}

Answer 2

"第一个成功的任务"的问题是如果所有任务都失败了怎么办？拥有永不完成的任务是一个非常糟糕的主意.

我假设你想要传播最后一个任务的异常,如果它们都失败了.考虑到这一点,我会说这样的事情是合适的:

async Task<Task<T>> FirstSuccessfulTask(IEnumerable<Task<T>> tasks)
{
  Task<T>[] ordered = tasks.OrderByCompletion();
  for (int i = 0; i != ordered.Length; ++i)
  {
    var task = ordered[i];
    try
    {
      await task.ConfigureAwait(false);
      return task;
    }
    catch
    {
      if (i == ordered.Length - 1)
        return task;
      continue;
    }
  }
  return null; // Never reached
}

该解决方案建立在OrderByCompletion扩展方法是一部分的我AsyncEx库 ; Jon Skeet和Stephen Toub也存在替代实施.

Answer 3

直接的解决方案是等待任何任务，检查它是否处于 RanToCompletion 状态，如果不是，则再次等待除已经完成的任务之外的任何任务。

async Task<TResult> WaitForFirstCompleted<TResult>( IEnumerable<Task<TResult>> tasks )
{
    var taskList = new List<Task<TResult>>( tasks );
    while ( taskList.Count > 0 )
    {
        Task<TResult> firstCompleted = await Task.WhenAny( taskList ).ConfigureAwait(false);
        if ( firstCompleted.Status == TaskStatus.RanToCompletion )
        {
            return firstCompleted.Result;
        }
        taskList.Remove( firstCompleted );
    }
    throw new InvalidOperationException( "No task completed successful" );
}