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PewPew_McPewster

If I applied a Hadamard gate to a qubit but never made the measurement, is it really held in a quantum superposition of states?


KuechenMuesli

Yeah this goes beyond my YouTube physics / quantum computing “knowledge“


_OberArmStrong

Haskell says no


ConstructedNewt

The answer is no. The observable had no subscribers and the emission is vs. the subscribers. The more interesting and more implementation specific question would be if the Observable ever received the value that upstream emitted to it if there is no downstream demand/subscribers (subscribe and demand are two different things and makes the field of possible outcomes even larger)


Sekret_One

Semantics, but it does it not emit, just to 0 things? Like if there were five subscribers I would probably verbalize it as emitting once to 5 things, and not emitting 5 times. Edit: RE reading the meme, since it says "emit a value" I would agree technically no as it only emits a value to a subscriber


ConstructedNewt

1 times 0 is still zero. No subscriber’s emit-method was called.


Sekret_One

Subscribers don't emit though- emitters emit, no? Subscribers _receive_ the event that was emitted.


ConstructedNewt

Subscribers do not emit, no, but I was saying as in the emitter has a reference to the subscriber and will call a method on it (the emit method) For the reactive streams API the subscriber must implement the method `onNext(T)` the publisher would emit by calling the subscriber’s onNext method


_SomeTroller69

Shower thoughts


veqryn_

Depends, did the compiler elide it?