Swift:确保不同协议与相关类型的等价性

我在一个项目中使用干净的Swift模式。该模式旨在解决MVC模式经常出现的大量视图控制器问题。

一般的想法是,ViewController将处理用户交互和一般业务逻辑的所有逻辑都推迟到交互器。该交互器执行其任务,然后将结果数据输出发送给演示者。演示者将数据包装到“视图模型”中,最后将其传递回ViewController。

这将使我能够轻松编写许多视图控制器需要的非常通用的函数(比如在加载时显示微调器、在safari视图中打开url等等)。

protocol CleanViewController: class {
    associatedType Interactor: CleanInteractor
    associatedType Router: CleanRouter
    init()
    init(interactor: Interactor, router: Router)
    var interactor: Interactor! { get set }
    var router: Router! { get set }
}

protocol CleanInteractor: class {
    associatedType Presenter: CleanPresenter
    init()
    init(presenter: Presenter)
    var presenter: Presenter! { get set }
}

protocol CleanPresenter: class {
    associatedType ViewController: CleanViewController
    init()
    init(viewController: ViewController)
    var viewController: ViewController? { get set }
}

protocol CleanRouter: class {
    associatedType DataStore: CleanDataStore
    associatedType ViewController: CleanViewController
    init()
    init(dataStore: DataStore, viewController: ViewController)
    var dataStore: DataStore! { get set }
    var viewController: ViewController? { get set }
}

所有这些都是相互参照的。这意味着,当我想要实现一个新的ViewController时,我需要以某种方式将它们链接在一起。现在这是通过定义 setup() 函数,从UIViewController的所有初始化器调用。

然后,每当我需要编写一个新的ViewController时,例如Login,它的实现方式如下:

protocol LoginDisplayLogic: CleanViewController {
    func display(_ some: ViewModel)
}

class LoginViewController<Interactor: LoginBusinessLogic, Router: LoginRoutingLogic>: LoginDisplayLogic, UIViewController {
    var interactor: Interactor!

    func display(_ some: ViewModel) { // display something }
    @IBAction func userDidTapLoginButton(_ sender: Any) { interactor.login() }
}

protocol LoginBusinessLogic: CleanInteractor {
    func login()
}

protocol LoginDataStore: CleanDataStore {
    var email: String { get set }
    var password: String { get set }
}

class LoginInteractor<Presenter: LoginPresentationLogic>: LoginBusinessLogic, LoginDataStore {
    var presenter: Presenter!

    var email: String
    var password: String

    func login() {
        // do something with email & password
    }
}

protocol LoginPresentationLogic: CleanPresenter {
    func display(_ loginResult: Result)
}

class LoginPresenter<ViewController: LoginDisplayLogic>: LoginPresentationLogic {
    var viewController: ViewController?

    func display(_ loginResult: Result) { 
        // wrap the result to a viewModel and call the viewController
    }
}

protocol LoginRoutingLogic: CleanRouter {
    func route(to some: Where)
}

class LoginRouter<DataStore: LoginDataStore, ViewController: LoginDisplayLogic>: LoginRoutingLogic {
   func route(to some: Where) { // ... }
}

现在,我要做这个 设置() 函数泛型。由于每个协议都定义了自己的关联类型,所有这些类型都需要一个初始化器,所以我只需要编写一次。

所以我写了以下内容:

extension CleanViewController where 
    Interactor: CleanDataStore, 
    Interactor.Presenter.ViewController == Self, 
    Router.ViewController == Self, 
    Router.DataStore == Interactor {

    func setup() {
        let presenter = Interactor.Presenter(viewController: self)
        let interactor = Interactor(presenter: presenter)
        let router = Router(viewController: self, dataStore: interactor)
        self.interactor = interactor
        self.router = router
    }
}

• Referencing instance method 'setup()' on 'CleanViewController' requires the types 'LoginViewController<Interactor, Router>' and 'Router.ViewController' be equivalent
• Referencing instance method 'setup()' on 'CleanViewController' requires the types 'Interactor' and 'Router.DataStore' be equivalent
• Referencing instance method 'setup()' on 'CleanViewController' requires the types 'Router.ViewController' and 'Interactor.Presenter.ViewController' be equivalent
• Referencing instance method 'setup()' on 'CleanViewController' requires the types 'Router' and 'Interactor.Presenter.ViewController.Router.ViewController.Router' be equivalent

这对我来说似乎很奇怪,因为延伸到哪里 设置() 声明的应确保所有这些关联类型都是等效的。

我知道这篇文章很长;复杂的一个,但我在这项工作中到底错过了什么?