异步和等待 - 难以理解

Question

我已尝试通读 Microsoft 的一些支持，但我试图将其简化，因为我发现其中大部分内容令人困惑。我不明白为什么要执行该方法的其余部分？看起来它一碰到await就跳出来然后再也没有回来，为什么它不回来并从它离开的地方开始呢？我试图让程序在咖啡等待时跳出，十个开始在鸡蛋上，所以让它们异步执行。任何帮助表示赞赏 - 谢谢。

 class Program
{
    static async Task Main(string[] args)
    { 
        Stopwatch stopWatch = new Stopwatch();
        Task<Coffee> coffeeTask = MakeCoffeeAsync(stopWatch);           
        Task<Egg> eggTask = FryEggs(2);
        stopWatch.Stop();
        TimeSpan ts = stopWatch.Elapsed;
        string elapsedTime = String.Format("{0:00}", ts.Seconds);
        Console.WriteLine("Time to make breakfast: " + elapsedTime + " seconds");
    }                

    private static async Task<Egg> FryEggs(int howMany)
    {
        Console.WriteLine("Warming the egg pan...");
        await Task.Delay(3000);
        Console.WriteLine($"cracking {howMany} eggs");
        Console.WriteLine("cooking the eggs ...");
     //  await Task.Delay(8000);
        Console.WriteLine("Put eggs on plate");
        Console.WriteLine("eggs are ready");
        return new Egg();
    }

    private static async Task<Coffee> MakeCoffeeAsync(Stopwatch timerstart)
    {
        timerstart.Start();
        Console.WriteLine("Turning on machine");
        Console.WriteLine("Waiting on machine to be ready");
        await Task.Delay(5000);
        Console.WriteLine("Getting cup");
       // await Task.Delay(4000);
        Console.WriteLine("weighing beans");
        Console.WriteLine("tamping coffee");
       // await Task.Delay(8000);
        Console.WriteLine("pouring coffee into cup");
        Console.WriteLine("coffee is ready" + "\n");
        return new Coffee();
    }
}
class Egg { }
class Coffee { }

这是输出：

Waiting on machine to be ready
Warming the egg pan...
Time to make breakfast: 00 seconds

Answer 1

您可以等待所有任务执行完毕：

await Task.WhenAll(eggTask, coffeeTask);

static async Task Main(string[] args)
{ 
    Stopwatch stopWatch = new Stopwatch();
    Task<Coffee> coffeeTask = MakeCoffeeAsync(stopWatch);           
    Task<Egg> eggTask = FryEggs(2);
    await Task.WhenAll(eggTask, coffeeTask);// NOTE THIS
    stopWatch.Stop();
    TimeSpan ts = stopWatch.Elapsed;
    string elapsedTime = String.Format("{0:00}", ts.Seconds);
    Console.WriteLine("Time to make breakfast: " + elapsedTime + " seconds");
} 

另一种选择是添加await关键字：

static async Task Main(string[] args)
{ 
    Stopwatch stopWatch = new Stopwatch();
    Coffee coffeeTask = await MakeCoffeeAsync(stopWatch);           
    Egg eggTask = await FryEggs(2);
    stopWatch.Stop();
    TimeSpan ts = stopWatch.Elapsed;
    string elapsedTime = String.Format("{0:00}", ts.Seconds);
    Console.WriteLine("Time to make breakfast: " + elapsedTime + " seconds");
} 

Answer 2

异步方法返回一个任务，该任务可以有不同的状态：任务可能仍在运行，也可能已完成。如果添加await，程序将等待任务完成。因为您没有await，所以您的程序不会等到早餐吃完。

static async Task Main(string[] args)
    { 
        Stopwatch stopWatch = new Stopwatch();
        Task<Coffee> coffeeTask = MakeCoffeeAsync(stopWatch);           
        Task<Egg> eggTask = FryEggs(2);
        Coffee cofee = await cofeeTask;
        Egg egg = await eggTask;
        stopWatch.Stop();
        TimeSpan ts = stopWatch.Elapsed;
        string elapsedTime = String.Format("{0:00}", ts.Seconds);
        Console.WriteLine("Time to make breakfast: " + elapsedTime + " seconds");
    } 

Answer 3

你正在解雇/忘记你的任务。您应该等待它们以避免在完成执行之前退出程序：

    static async Task Main(string[] args)
    { 
        Stopwatch stopWatch = new Stopwatch();
        Task<Coffee> coffeeTask = MakeCoffeeAsync(stopWatch);           
        Task<Egg> eggTask = FryEggs(2);
        Coffee coffee = await coffeeTask;
        Egg egg = await eggTask;
        stopWatch.Stop();
        TimeSpan ts = stopWatch.Elapsed;
        string elapsedTime = String.Format("{0:00}", ts.Seconds);
        Console.WriteLine("Time to make breakfast: " + elapsedTime + " seconds");
    }

Answer 4

您的流程永远不会返回的原因是因为您没有等待函数 FryEggs 和 MakeCoffeeAsync。

await 是一个关键字，它告诉您的代码在这里停止，直到它有结果。

如果你想同时煮咖啡和鸡蛋，你想awaitFryEggs。

如果你想煮咖啡，那就煮鸡蛋，你想awaitFryEggs 和MakeCoffeeAsync 都可以