使用Func委托与Async方法

Question

我试图使用Func与异步方法.我收到了一个错误.

无法将异步lambda表达式转换为委托类型'Func<HttpResponseMesage>'.异步lambda表达式可能返回void,Task或者Task<T>都不能转换为'Func<HttpResponseMesage>'.

以下是我的代码:

public async Task<HttpResponseMessage> CallAsyncMethod()
{
    Console.WriteLine("Calling Youtube");
    HttpClient client = new HttpClient();
    var response = await client.GetAsync("https://www.youtube.com/watch?v=_OBlgSz8sSM");
    Console.WriteLine("Got Response from youtube");
    return response;
}

static void Main(string[] args)
{
    Program p = new Program();
    Task<HttpResponseMessage> myTask = p.CallAsyncMethod();
    Func<HttpResponseMessage> myFun =async () => await myTask;
    Console.ReadLine();
}

Answer 1

正如错误所说,异步方法返回Task,Task<T>或void.为了让这个工作,您可以:

Func<Task<HttpResponseMessage>> myFun = async () => await myTask;

Answer 2

我通常采取的路径是让该Main方法调用Run()返回任务的方法，然后.Wait()完成Task。

class Program
{
    public static async Task<HttpResponseMessage> CallAsyncMethod()
    {
        Console.WriteLine("Calling Youtube");
        HttpClient client = new HttpClient();
        var response = await client.GetAsync("https://www.youtube.com/watch?v=_OBlgSz8sSM");
        Console.WriteLine("Got Response from youtube");
        return response;
    }

    private static async Task Run()
    {
        HttpResponseMessage response = await CallAsyncMethod();
        Console.ReadLine();
    }

    static void Main(string[] args)
    {
        Run().Wait();
    }
}

这允许控制台应用程序的其余部分在完全异步/等待支持的情况下运行。由于控制台应用程序中没有任何 UI 线程，因此使用.Wait().