Why is Rxjs publishReplay (1) .refCount () not playing?

Question

Why does publishReplay (1) .refCount () not reproduce the last value for late subscribers?

a = new Rx.Subject(); b = a.publishReplay(1).refCount(); a.subscribe(function(x){console.log('timely subscriber:',x)}); a.next(1); b.subscribe(function(x){console.log('late subscriber:',x)}); 

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Expected Result:

 timely subscribe: 1 late subscriber: 1 

Actual output

 timely subscriber: 1 

Answer 1

This is because at the time you call a.next(1) , publishReplay(1) did not subscribe to its Observable source (Subject a in this case), and therefore the internal ReplaySubject will not get the value 1 .

In RxJS 5, the actual subscription between operators occurs when you sign at the end of the chain, which is b.subscribe(...) in this example. Cm:

• https://github.com/ReactiveX/rxjs/blob/master/src/operator/multicast.ts#L63
• https://github.com/ReactiveX/rxjs/blob/master/src/Observable.ts#L96

Until you name subscribe() , the operators are bound using the lift() method, which simply takes an instance of the operator and assigns it to a new Observable. The operator.call() method, as you can see in the two links above, is called later when subscribing. Cm:

• https://github.com/ReactiveX/rxjs/blob/master/src/Observable.ts#L67

Answer 2

Your first subscriber subscribes to a , since refCount first activates the stream, if there is at least one subscriber (who is not, because not b , but a signed), this is not active until your last loc.

 a = new Rx.Subject(); b = a.publishReplay(1).refCount(); b.subscribe(function(x){console.log('timely subscriber:',x)}); a.next(1); b.subscribe(function(x){console.log('late subscriber:',x)}); 

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