我已经阅读了Rx
Swift / ShareReplayScope.swift文件,但有点难以理解.
public func share(replay: Int = 0,scope: SubjectLifetimeScope = .whileConnected) -> Observable<E> { switch scope { case .forever: switch replay { case 0: return self.multicast(PublishSubject()).refCount() default: return self.multicast(ReplaySubject.create(bufferSize: replay)).refCount() } case .whileConnected: switch replay { case 0: return ShareWhileConnected(source: self.asObservable()) case 1: return ShareReplay1WhileConnected(source: self.asObservable()) default: return self.multicast(makeSubject: { ReplaySubject.create(bufferSize: replay) }).refCount() } } }
0,1和默认值,有什么区别?为什么将1与defalut分开?
override func subscribe<O : ObserverType>(_ observer: O) -> disposable where O.E == E { _lock.lock() let connection = _synchronized_subscribe(observer) let count = connection._observers.count let disposable = connection._synchronized_subscribe(observer) _lock.unlock() if count == 0 { connection.connect() } return disposable }
如何锁定工作,最困难的是这个功能.封锁的obserables如何正确连接到他们的观察者.
解决方法
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