c# - 使用 'yield return' 观察异步序列

Question

以下示例工作正常:

static IEnumerable<int> GenerateNum(int sequenceLength)
    {
      for(int i = 0; i < sequenceLength; i++)
      {
          yield return i;
      }
    }

static void Main(string[] args)
    {

        //var observ = Observable.Start(() => GenerateNum(1000));
        var observ = GenerateNum(1000).ToObservable();

        observ.Subscribe(
            (x) => Console.WriteLine("test:" + x),
            (Exception ex) => Console.WriteLine("Error received from source: {0}.", ex.Message),
            () => Console.WriteLine("End of sequence.")
            );

        Console.ReadKey();
    }

但是，我真正想要的是使用注释掉的行——即我想异步运行“数字生成器”，每次它产生一个新值时，我都希望将它输出到控制台。它似乎不起作用 - 我如何修改此代码才能起作用？

Answer 1

在控制台应用程序中为异步执行执行此操作时，您可能需要使用 ToObservable(IEnumerable<TSource>, IScheduler)过载(参见 Observable.ToObservable Method (IEnumerable, IScheduler) )。例如，要使用内置的线程池调度，请尝试

var observ = GenerateNum(1000).ToObservable(Scheduler.ThreadPool);

它对我有用...为了扩展，下面的完整示例完全按照我认为您的意图工作:

    static Random r = new Random();

    static void Main(string[] args) {

        var observ = GenerateNum(1000).ToObservable(Scheduler.ThreadPool );

        observ.Subscribe(
            (x) => Console.WriteLine("test:" + x),
            (Exception ex) => Console.WriteLine("Error received from source: {0}.", ex.Message),
            () => Console.WriteLine("End of sequence.")
            );

        while (Console.ReadKey(true).Key != ConsoleKey.Escape) {
            Console.WriteLine("You pressed a key.");
        }
    } 

    static IEnumerable<int> GenerateNum(int sequenceLength) {
        for (int i = 0; i < sequenceLength; i++) {
            Thread.Sleep(r.Next(1, 200));
            yield return i;
        }
    }