如何将 currentValueSubject 转换为不可变的 observable？

Question

所以我RxSwift比更熟悉Combine。管理可变/不可变接口的一个好方法是我在RxSwift

protocol SampleStream {
   /// An immutable interface. 
   var streamInfo: Observable<String?> { get} 
}

protocol MutableSampleStream: SampleStream {
   /// A mutable interface. 
   func updateStream( _ val: String?)
}

func SampleStreamImpl: MutableSampleStream {

   // Returns the immutable version of the stream.
   // If I pass down SampleStream as a dependency, then nothing else can write to this stream.
   // When they subscribe, they immediately get a value though since it's a behavior subject. 
   var streamInfo: Observable<String?> {
      return streamInfoSubject.asObservable()
   }

   private var streamInfoSubject = BehaviorSubject<String?>(value: nil) 

   func updateStream { }
}

我怎样才能使用 做类似的事情Combine？组合currentValueSubject似乎没有办法将其转换为非读写版本。或者我错过了什么？

在我的应用程序中，我不想直接传递 acurrentValueSubject因为我知道我只希望从一个地方更新此流。其他地方应该只从流中读取，而不具有写入功能。

Answer 1

用作AnyPublisher您的非可变类型：

protocol SampleStream {
    var streamInfo: AnyPublisher<String?, Error> { get }
}

protocol MutableSampleStream: SampleStream {
    func updateStream(_ val: String?)
}

class MySampleStream: MutableSampleStream {
    var streamInfo: AnyPublisher<String?, Error> {
         return subject.eraseToAnyPublisher()
    }

    func updateStream(_ val: String?) { subject.send(val) }

    private let subject = CurrentValueSubject<String?, Error>(nil)
}