Rxswift observable bind(to :) vs subscribe(onNext :)

概述抱歉.我很困惑Rx swift中的绑定是什么.据我所知,除非观察者订阅了它,否则observable不会产生价值,例如myObservable.subscribe(onNext：{}). 但是,当我阅读以下代码行时： // in LoginViewModel.swiftinit() { isValid = Observable.combineLatest(username.asObser 抱歉.我很困惑Rx swift中的绑定是什么.据我所知,除非观察者订阅了它,否则observable不会产生价值,例如myObservable.subscribe(onNext：{}).
但是,当我阅读以下代码行时：

// in Loginviewmodel.swiftinit() {    isValID = Observable.combineLatest(username.asObservable(),password.asObservable()) { (username,password) in        return !username.isEmpty && !password.isEmpty    }}// in LoginVIEwController.swiftviewmodel.isValID.bind(to: loginbutton.rx.isEnabled).disposed(by: disposeBag)

我很困惑,为什么在不调用subscribe方法的情况下能够观察到isValID Observable？为什么我们可以在LoginVIEwController.swift中调用bind(to :)而不调用viewmodel.isValID.subscribe(…)之类的东西

看看bind(to :)的实现

public func bind<O: ObserverType>(to observer: O) -> disposable where O.E == E {    return self.subscribe(observer)}

订阅在里面调用.

关于你的陈述

As far as I kNow,observable won’t produce value unless a observer subscribed on it

这只适用于寒冷的观察者.让我引用这个site

When does an Observable begin emitting its sequence of items? It depends on the Observable. A “hot” Observable may begin emitting items as soon as it is created,and so any observer who later subscribes to that Observable may start observing the sequence somewhere in the mIDdle. A “cold” Observable,on the other hand,waits until an observer subscribes to it before it begins to emit items,and so such an observer is guaranteed to see the whole sequence from the beginning.

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