Geek asks and answers

Why can't IObservable <T> be used without the Reactive (Rx) extension for .NET?

Trying to understand why and when I will need to use Reactive Extension (Rx) for .NET, I came up with the question " Cyno aync / awue C # 5.0 and Rx - Reactive Extensions ", which, by reading through links, led to more questions than responses.

Link provided in the last article Task versus IObservable: when to use what? seems to reference IObservablesynonymously (or interchangeably) using the Reactive (Rx) extension for .NET

What, for example, the phrase from the specified article:

Your code will require reactive extensions if you decide to return IObservable

wanted to say?

Both Task<T>and IObservable<T>are part of .NET, I use without any reference or Rx settings.
Why do I require reactive extensions in order to retuen IObservable?

What does RX do in the discussion Task<T>against IObservable?
And why are their use comparable?

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2 answers

In fact, this is due to the ease of implementation. You should never try to implement on your IObservable<T>own; Rx has any implementation that you could do correctly and efficiently.

Rx IObservable<T>. , MSDN .

Update:

MSDN ( , ):

  • Rx (4.1).
  • OnError (4.3).
  • (6.1, 6.2).
  • OnError (6.6).
  • , (6.17).

, , TrackLocation . Rx .

, Rx Synchronize, TrackLocation.

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IObservable IObserver Rx Rx... .NET Framework, IObservables , Rx, - IObservable, Rx- , Rx.

IObservable IObserver ( "", , , ) IEnumerable IEnumerator. , IEnumerable MoveNext Current foreach, ( , ), LINQ , IE, Rx -. , BCL , , , , foreach, , , , IO.

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