相关疑难解决方法(0)

如果我不关心任务完成的顺序,只需要完成它们,我还应该使用await Task.WhenAll而不是多个await吗？例如,DoWork2低于优选的方法DoWork1(以及为什么？):

using System;
using System.Threading.Tasks;

namespace ConsoleApp
{
    class Program
    {
        static async Task<string> DoTaskAsync(string name, int timeout)
        {
            var start = DateTime.Now;
            Console.WriteLine("Enter {0}, {1}", name, timeout);
            await Task.Delay(timeout);
            Console.WriteLine("Exit {0}, {1}", name, (DateTime.Now - start).TotalMilliseconds);
            return name;
        }

        static async Task DoWork1()
        {
            var t1 = DoTaskAsync("t1.1", 3000);
            var t2 = DoTaskAsync("t1.2", 2000);
            var t3 = DoTaskAsync("t1.3", 1000);

            await t1; await t2; await t3;

            Console.WriteLine("DoWork1 results: {0}", String.Join(", ", …

我正在阅读Stephen Cleary撰写的"C#Cookbook中的并发",我注意到以下技巧:

var completedTask = await Task.WhenAny(downloadTask, timeoutTask);  
if (completedTask == timeoutTask)  
  return null;  
return await downloadTask;  

downloadTask是对httpclient.GetStringAsync的调用,timeoutTask正在执行Task.Delay.

如果它没有超时,则downloadTask已经完成.为什么有必要做第二次等待而不是返回downloadTask.Result,因为任务已经完成了？

我有一个异步方法,GetExpensiveThing()它执行一些昂贵的I/O工作.这就是我使用它的方式:

// Serial execution
public async Task<List<Thing>> GetThings()
{
    var first = await GetExpensiveThing();
    var second = await GetExpensiveThing();
    return new List<Thing>() { first, second };
}

但由于这是一种昂贵的方法,我想并行执行这些调用.我本以为移动等待会解决这个问题:

// Serial execution
public async Task<List<Thing>> GetThings()
{
    var first = GetExpensiveThing();
    var second = GetExpensiveThing();
    return new List<Thing>() { await first, await second };
}

这不起作用,所以我将它们包装在一些任务中,这有效:

// Parallel execution
public async Task<List<Thing>> GetThings()
{
    var first = Task.Run(() =>
    {
        return GetExpensiveThing();
    });

    var second = Task.Run(() =>
    { …