Swift Combine:这些多播功能有什么作用，我该如何使用它们? [英] Swift Combine: What are those multicast functions for and how do I use them?

问题描述

在遇到一些合并问题时，我在 https:中遇到了使用多个订户"部分: //developer.apple.com/documentation/combine/publisher :

Struggling with some combine problems I came across the "Working with Multiple Subscribers" section in https://developer.apple.com/documentation/combine/publisher :

func multicast<S>(() -> S) -> Publishers.Multicast<Self, S>

func multicast<S>(subject: S) -> Publishers.Multicast<Self, S>

但是，当我尝试确认我的假设，即发送给多个订阅者时将需要多播时，我发现在尝试使用此操场代码(从

However, when I tried to confirm my assumption that multicast would be needed when sending to multiple subscribers, I found out this is not necessary when trying on this playground code (modified from https://github.com/AvdLee/CombineSwiftPlayground/blob/master/Combine.playground/Pages/Combining%20Publishers.xcplaygroundpage/Contents.swift ) (run on 10.14.5 in Xcode Version 11.0 beta 3 (11M362v)):

enum FormError: Error { }

let usernamePublisher = PassthroughSubject<String, FormError>()
let passwordPublisher = PassthroughSubject<String, FormError>()

let validatedCredentials = Publishers.CombineLatest(usernamePublisher, passwordPublisher)
    .map { (username, password) -> (String, String) in
        return (username, password)
    }
    .map { (username, password) -> Bool in
        !username.isEmpty && !password.isEmpty && password.count > 12
    }
    .eraseToAnyPublisher()

let firstSubscriber = validatedCredentials.sink { (valid) in
    print("First Subscriber: CombineLatest: Are the credentials valid: \(valid)")
}

let secondSubscriber = validatedCredentials.sink { (valid) in
    print("Second Subscriber: CombineLatest: Are the credentials valid: \(valid)")
}

// Nothing will be printed yet as `CombineLatest` requires both publishers to have send at least one value.
usernamePublisher.send("avanderlee")
passwordPublisher.send("weakpass")
passwordPublisher.send("verystrongpassword")

此打印:

First Subscriber: CombineLatest: Are the credentials valid: false
Second Subscriber: CombineLatest: Are the credentials valid: false
First Subscriber: CombineLatest: Are the credentials valid: true
Second Subscriber: CombineLatest: Are the credentials valid: true

因此，似乎不需要多播即可寻址多个订户.还是我错了?

so it seems that no multicast is needed to address multiple subscribers. Or I am wrong?

那么，这些多播功能有什么作用，我将如何使用它们?一些示例代码会很好.

So, what are those multicast functions for and how would I use them? Some example code would be nice.

谢谢

Lars

推荐答案

PassthroughSubject并不是一个很好的示例，因为它是一个类，并且为您提供了参考语义.因此，在一个简单的情况下，每当主题发出一个主题时，两个订阅者就可以直接订阅它并同时接收相同的值.

PassthroughSubject is not a very good example to test with, because it's a class and gives you reference semantics. Therefore in a simple case two subscribers can subscribe directly to it and receive the same values at the same time whenever the subject emits one.

但这是一个更好的测试用例(灵感来自可可与爱(a)):

But here's a better test case (inspired by a discussion on Cocoa With Love):

    let pub1 = Timer.publish(every: 1, on: .main, in: .default).autoconnect()
    let sub = CurrentValueSubject<Int,Never>(0)
    let scan = sub.scan(10) {i,j in i+j}
    pub1.sink { _ in let i = sub.value; sub.value = i+1 }.store(in:&storage)
    scan.sink { print("a", $0) }.store(in:&storage)
    delay(3) {
        scan.sink { print("b", $0) }.store(in:&self.storage)
    }

当第二个sink作为该管道的新订阅者出现时，得出的结果肯定很奇怪:

That gives a decidedly weird result when the second sink comes along as a new subscriber to this pipeline:

a 10
a 11
a 13
a 16
b 13
a 20
b 17
a 25
b 22
a 31
b 28
a 38
b 35

接收器a和b正在获得彼此不同的一系列数字，这实际上是因为scan是一个结构.如果我们希望他们获得相同的数字，则可以使用多播:

Sinks a and b are getting a different series of numbers from one another, effectively because scan is a struct. If we want them to get the same numbers, we can use a multicast:

    let scan = sub.scan(10) {i,j in i+j}.multicast {PassthroughSubject()}.autoconnect()

这产生了

a 10
a 11
a 13
a 16
a 20
b 20
a 25
b 25

这是连贯的.

但是， still 并不能证明您需要multicast，因为您可以通过说.share()来完成相同的事情.我不清楚multicast和share之间的区别是什么.

However, that still doesn't prove you need multicast, because you could accomplish the same thing by saying .share() instead. I'm not clear on what the difference is between multicast and share.

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