Question

我从反应框架开始..基于示例：

http://rxwiki.wikidot.com/101samples（乒乓球）

我在我的应用程序中执行了类似的实现..我不能也不知道如何将操作传递给OnError（在类之外访问，ISubject实现）

我该怎么办？

public class PFocusPing : ISubject<PFocusPong, PFocusPing>
{
    private readonly AdvancedConnection device;
    public PFocusPing(AdvancedConnection con)
    {
        this.device = con;
    }

    #region Implementation of IObserver<PFocusPong>
    public void OnNext(PFocusPong value)
    {
        this.device.SendData(clsPowerFocus.DataKeepAlive);
        Console.WriteLine("PFocusPing received PFocusPong.");
    }

    public void OnError(Exception exception)
    {
        Console.WriteLine("PFocusPing experienced an exception and had to quit playing.");
    }

    public void OnCompleted()
    {
        Console.WriteLine("PFocusPing finished.");
    }
    #endregion

    #region Implementation of IObservable<PFocusPing>
    public IDisposable Subscribe(IObserver<PFocusPing> observer)
    {
        return Observable.FromEventPattern<Args<DataReceive>>(add => device.DataReceived += add,
                                                     rem => device.DataReceived -= rem)
                                                    .Where(obj => obj.EventArgs.Value.Result == DataReceive.CONNECTION_RESULT.SUCESS)
                                                    .Where(obj => obj.EventArgs != null &&
                                                                  obj.EventArgs.Value != null &&
                                                                  obj.EventArgs.Value.Data != null &&
                                                                  obj.EventArgs.Value.Data.Length > 0)
                                                    .Select(obj => this)
                                                    .Subscribe(observer);

    }
    #endregion

    #region Implementation of IDisposable
    public void Dispose()
    {
        OnCompleted();
    }
    #endregion
 }

public class PFocusPong : ISubject<PFocusPing, PFocusPong>
{
    private readonly int timeout;
    private readonly int reply;
    private DateTime lastAnswer;

    public PFocusPong(int msPing, int msTimeout)
    {
        this.reply = Math.Max(1000, msPing);
        this.timeout = Math.Max(2000, msTimeout);
    }

    #region Implementation of IObserver<Ping>

    public void OnNext(PFocusPing value)
    {
        this.lastAnswer = DateTime.Now;
        Console.WriteLine("PFocusPong received Ping.");
    }

    public void OnError(Exception exception)
    {
        Console.WriteLine("PFocusPong experienced an exception and had to quit playing.");
    }

    public void OnCompleted()
    {
        Console.WriteLine("PFocusPong finished.");
    }

    #endregion

    #region Implementation of IObservable<PFocusPong>

    public IDisposable Subscribe(IObserver<PFocusPong> observer)
    {
        var keepAlive =
            Observable.Timer(TimeSpan.FromMilliseconds(0), TimeSpan.FromMilliseconds(reply))
            .Select(obj =>
                {
                    if ((DateTime.Now - this.lastAnswer).TotalMilliseconds < reply)
                    {
                        Console.WriteLine("True");
                        return true;
                    }

                    Console.WriteLine("False");
                    return false;
                });

        var keepDead = keepAlive.Where(obj => obj).Timeout(TimeSpan.FromMilliseconds(timeout)).Subscribe(r =>
            {
                if (!r)
                    throw new TimeoutException();
            });


        var disposable = keepAlive.Finally(keepDead.Dispose).Select(n => this).Subscribe(observer);
        return disposable;
    }

    #endregion

    #region Implementation of IDisposable

    public void Dispose()
    {
        OnCompleted();
    }

    #endregion

}

Answer 1

我不确定你的问题是什么。但是，您的困惑可能来自您的起点。我强烈建议您不要实施ISubject<T1, T2>，ISubject<T>，IObserver<T>或IObservable<T>接口。

您应该使用Subscribe(Action<T>, Action<Exception>, Action)扩展方法订阅可观察序列（IObservable<T>），并使用Observable.Create工厂方法或其他转换方法（例如FromEvent，ToObservable等）。。）

同样乒乓球并不是真正的Rx精神。 Rx是关于观察和组合事件序列的。乒乓似乎更像是传递信息。 Rx通常是一个方向的东西，即我会听键盘事件并做某事，或听取温度变化，新价格等......

当你收到活动时，你通常不会把钥匙推回去。

捕获异常（乒乓反应样本）

1 个答案: