如何正确等待第一个完成的任务

Question

我正在编写一个简单的控制台应用程序，该应用程序应该向不同的外部源发出请求，并且 return 来自首先响应的源的响应。

我正在尝试使用 Task 对象来执行此操作（以使代码尽可能简单），但我的程序死锁并且永远不会完成执行。现在，我怀疑这是因为我在我的任务中使用 Result，但我不知道如何在没有 Result.

的情况下使其工作

public static string GetResultFromAnySource()
{
    var t1 = new Task<string>(() => { Thread.Sleep(500); return "Result from source 1"; });
    var t2 = new Task<string>(() => { Thread.Sleep(100); return "Result from source 2"; });

    var res = Task.WhenAny(t1, t2).Result;

    if (res == t1)
        return "Source 1 was faster. Result: " + t1.Result;
    else if (res == t2)
        return "Source 2 was faster. Result: " + t2.Result;

    throw new ApplicationException("Something went very wrong");
}

static void Main(string[] args)
{
    Console.WriteLine(GetResultFromAnySource());
}

感谢任何帮助。

Answer 1

您错过了开始任务的时间。

t1.Start();
t2.Start();

或者，使用 Task.Run()

var t1 = Task.Run<string>(() => { Thread.Sleep(500); return "Result from source 1"; });
var t2 = Task.Run<string>(() => { Thread.Sleep(100); return "Result from source 2"; });

Answer 2

嗨，安德烈，我使用 async/await 个关键字重写了您的程序：

class Program {
public static async Task<string> GetResultFromAnySource() {

  var t1 = Task<string>.Run( () => {  Task.Delay(500).Wait(); return "Result from source 1"; });
  var t2 = Task<string>.Run(() => { Task.Delay(100).Wait(); return "Result from source 2"; });

  var res = await Task.WhenAny(t1, t2);

  if (res == t1)
    return "Source 1 was faster. Result: " + await t1;
  else if (res == t2)
    return "Source 2 was faster. Result: " + await t2;

  throw new ApplicationException("Something went very wrong");
}

static void Main(string[] args) {

  Task.Run(async () => await Out()).Wait();
}

static async Task Out()
{
  var str = await GetResultFromAnySource();
  Console.WriteLine(str);
}

}

推荐使用Task.Run-代码清晰，立即启动任务。

如何正确等待第一个完成的任务

How to properly wait for first completed task

c#

console-application

task-parallel-library