c# – 使用Task的TaskCanceledException

我正在做的事情的概述：在一个循环中,我正在开始一个新的任务< string>并将其添加到列表< Task< string>>.问题是,在返回字符串后,任务抛出System.Threading.Tasks.TaskCanceledException,我不知道为什么.下面是我正在做的修剪版本

public async Task<string> GenerateXml(object item)
{
    using (var dbContext = new DatabaseContext())
    {
        //...do some EF dbContext async calls here
        //...generate the xml string and return it
        return "my xml data";
    }
}

var tasks = new List<Task<string>>();

我的循环看起来像：

foreach (var item in items)
{
    tasks.Add(Task.Run(() => GenerateXml(item).ContinueWith((t) => { return ""; },TaskContinuationOptions.OnlyOnFaulted)));
    //also tried: 
    tasks.Add(Task.Run(async () => await GenerateXml(item).ContinueWith((t) => { return ""; },TaskContinuationOptions.OnlyOnFaulted)));
    //both generate the same exception

    //after looking at my code,I was using the ContinueWith on the GenerateXml method call,which should still work,right?
    //I moved the continue with to the `Task.Run` and still get the exception.
}

Task.WaitAll(tasks.ToArray()); //this throws the AggregateException which contains the TaskCanceledException

当我单步执行代码时,它会返回“my xml data”;但抛出异常.

我想用ContinueWith避免的是当我循环每个任务并获得结果时,它不会抛出与WaitAll抛出的相同的AggregateException.

这是一个工作的控制台应用程序抛出…我知道问题是与ContinueWith,但为什么？

class Program
{
    static void Main(string[] args)
    {
        var program = new Program();
        var tasks = new List<Task<string>>();

        tasks.Add(Task.Run(() => program.GenerateXml().ContinueWith((t) => { return ""; },TaskContinuationOptions.OnlyOnFaulted)));

        Task.WaitAll(tasks.ToArray()); //this throws the AggregateException

        foreach (var task in tasks)
        {
            Console.WriteLine(task.Result);
        }

        Console.WriteLine("finished");
        Console.ReadKey();
    }

    public async Task<string> GenerateXml()
    {
        System.Threading.Thread.Sleep(3000);
        return "my xml data";
    }
}

Task<string> t1 = Task.Run(() => program.GenerateXml());
    t1.ContinueWith((t) => { return ""; },TaskContinuationOptions.OnlyOnFaulted);
    tasks.Add(t1);

tasks.Add(program.GenerateXml().ContinueWith(t => {return t.IsFaulted? "": t.Result; }));