swift - 如何通过 Swift 3 中的协议(protocol)从类公开 init() 的接口(interface)

Question

我想通过 swift 协议(protocol)在类中使用泛型类型:

public protocol WebSocketType {
    // some interfaces
}

class _Network<WebSocketT: WebSocketType, Configuration: NetworkConfigurationType> {
    // Use the generic type
    let websocket: WebSocketT
    init(host: String, api: String) {
        // do something here...

        // create the instance which will conform the protocol via generic type
        // compilation error: 'WebSocketT' cannot be constructed because it has no accessible initializers
        self.websocket = WebSocketT() 
    }
}

// I'll create a class with concrete classes (WebSocket and NetworkConfiguration). 
// I don't have the class `WebSocket` and I'll extend that to conform the protocol `WebSocketType`.
extension WebSocket: WebSocketType {}
typealias Network = _Network<WebSocket, NetworkConfiguration>
let network = Network()

我收到错误 'WebSocketT' cannot be constructed because it has no accessible initializers ，我添加了 init()在协议(protocol)中WebSocketType :

public protocol WebSocketType {
    init()
}

然后我又遇到了一个错误 Initializer requirement 'init()' can only be satisfied by a必填initializer in non-final class 'WebSocket' 。

如何解决此问题以在类中构造泛型类型？

Answer 1

WebSocket 的子类不必继承所有 WebSocket 的初始化程序。因此，WebSocket 的子类可能无法符合 WebSocketType。解决方案是使用 final class WebSocket 声明 WebSocket，以便它不能被子类化，或者将 WebSocket 的初始化器声明为 required init () 以便所有子类也必须提供它。然后，WebSocket 将符合 WebSocketType，从而解决您的问题。