Is Rx.Subject a hot observable?

Question

The code

const a = new Rx.Subject().do(x => console.log(\'a\'))
const b = a.mapTo(0)
const c = a.mapTo(1)
const d = Rx.Observable.merge(b, c)
d.subscribe(         


        

Answer 1

The Subject itself is hot/shared.

However: Any(most!) operators that you append will create a new stream, with the previous stream(in this case the Subject) as source - the new stream, however, is (for most operators) not hot and will only be made hot by deriving a hot stream through appending a hot operator (like share or publish ect...)

So when you share your do, everything should work as expected.

const a = new Rx.Subject().do(x => console.log('a')).share();
const b = a.mapTo(0);
const c = a.mapTo(1);
const d = Rx.Observable.merge(b, c)
d.subscribe(x => console.log('d'));
a.next(3);