如何在C#中观察由于另一个等待的任务失败而没有等待的任务？

Question

我有这样的代码：

var task1 = operation1();
var task2 = operation2();


var result1 = await task1;
var result2 = await task2;

附言。这个例子是简化的，我有多个任务运行这样。因此，我正在寻找一个通用的解决方案，将为任何数量的任务工作

Answer 1

作为task.whenall的替代方案，您可以对task.whenany使用渐进式方法。

当任何任务完成时，您可以检查结果，并决定下一步要做什么。（请记住任务。当任何都没有抛出异常时，即使它是等待的）这种方法的优点是您可以轻松地添加节流（控制并行度）。

static async Task ProgressiveAsyncForEach(int degreeOfParallelism, params Task[] tasks)
{
    var toBeProcessedTasks = new HashSet<Task>();
    var remainingTasksEnumerator = tasks.GetEnumerator();

    void AddNextTask()
    {
        if (!remainingTasksEnumerator.MoveNext()) return;
        var nextTaskToProcess = (Task)remainingTasksEnumerator.Current;
        toBeProcessedTasks.Add(nextTaskToProcess);
    }

    //Initialize
    while (toBeProcessedTasks.Count < degreeOfParallelism)
    {
        AddNextTask();
    }

    while (toBeProcessedTasks.Count > 0)
    {
        var processedTask = await Task.WhenAny(toBeProcessedTasks).ConfigureAwait(false);
        if (!processedTask.IsCompletedSuccessfully)
        {
            Console.WriteLine("One of the task has failed");
            //TODO: log first failed task
            //CONSIDER: cancel all the remaining tasks
            return;
        }

        toBeProcessedTasks.Remove(processedTask);
        AddNextTask();
    }
}

static async Task Main(string[] args)
{
    await ProgressiveAsyncForEach(2, Faulty(), Fast(), Slow());
    Console.WriteLine("Application finished");
}

static async Task Slow()
{
    Console.WriteLine("Slow started");
    await Task.Delay(1000);
    Console.WriteLine("Slow finished");
}

static async Task Fast()
{
    Console.WriteLine("Fast started");
    await Task.Delay(500);
    Console.WriteLine("Fast finished");
}

static async Task Faulty()
{
    Console.WriteLine("Faulty started");
    await Task.Delay(700);
    throw new Exception();
}