我如何 return 实现另一个协议的通用类型

Question

我有一个网络模块，它实现了一个标准接口来公开我的网络客户端

protocol HTTPClientTask {
    func cancel()
}
protocol HTTPClient {
    @discardableResult
    func execute(_ request: URLRequest, _ completion: @escaping (Result<(response: HTTPURLResponse, data: Data), Error>) -> Void) -> HTTPClientTask
}

这是像

这样实现的

final class URLSessionHTTPClient: HTTPClient {
    
    private let session: URLSession
    private struct RequestError: Error { }
    private struct URLSessionTaskWrapper: HTTPClientTask {
        let wrapped: URLSessionTask
        func cancel() {
            wrapped.cancel()
        }
    }
    
    init(session: URLSession = .shared) {
        self.session = session
    }
    
    func execute(_ request: URLRequest, _ completion: @escaping (Result<(response: HTTPURLResponse, data: Data), Error>) -> Void) -> HTTPClientTask {
        let task = session.dataTask(with: request) { data, response, error in
            completion(Result {
                if let error = error {
                    throw error
                } else if let data = data, let response = response as? HTTPURLResponse {
                    return (response, data)
                } else {
                    throw RequestError()
                }
            })
        }
        task.resume()
        return URLSessionTaskWrapper(wrapped: task)
    }
}

操场上运行的一个例子是

let requestURL = URL(string: "https://jsonplaceholder.typicode.com/todos/1")!

let httpClient = URLSessionHTTPClient()

httpClient.execute(.init(url: requestURL)) { result in
    if let code = try? result.get().response.statusCode {
        print(code)
    }
}

我有另一个框架我想添加到我的应用程序中，它公开了一个相同的界面。

我不希望我的其他模块依赖于这个网络模块，相反我更希望它公开它需要的接口并让我的网络模块依赖它。

所以下面的接口是我的其他模块暴露的

protocol AuthHTTPClientTask {
    func cancel()
}

protocol AuthHTTPClient {
    @discardableResult
    func execute(_ request: URLRequest, _ completion: @escaping (Result<(response: HTTPURLResponse, data: Data), Error>) -> Void) -> AuthHTTPClientTask
}

如您所见，接口是准确的，但是我希望避免只为该接口创建一个完整的 URLSessionHTTPClient - URLSessionAuthHTTPClient 或行为等相同的东西。

是否可以创建某种 URLSessionHTTPClient 实现的类型，使其能够 return AuthHTTPClientTask 或 HTTPClientTask？

我想我可以做类似的事情

extension HTTPClientTask: AuthHTTPClientTask { }

final class URLSessionHTTPClient: HTTPClient, AuthHTTPClient {
    
    private let session: URLSession
    private struct RequestError: Error { 
.........

但这会产生 Extension of protocol 'HTTPClientTask' cannot have an inheritance clause 和 Type 'URLSessionHTTPClient' does not conform to protocol 'AuthHTTPClient'

Answer 1

您收到有关 inheritance clause 的错误，因为出于一致性目的需要约束，在您的示例中，编译器认为这是继承。

您可以创建一个通用的私有方法来处理请求的分派，并扩展该方法的 return 类型以符合您的模块类型。

protocol HTTPClientTask {
    func cancel()
}

protocol HTTPClient {
    @discardableResult
    func execute(_ request: URLRequest, _ completion: @escaping (Result<(response: HTTPURLResponse, data: Data), Error>) -> Void) -> HTTPClientTask
}

protocol AuthHTTPClientTask {
    func cancel()
}

protocol AuthHTTPClient {
    @discardableResult
    func execute(_ request: URLRequest, _ completion: @escaping (Result<(response: HTTPURLResponse, data: Data), Error>) -> Void) -> AuthHTTPClientTask
}


private struct URLSessionTaskWrapper {
    let wrapped: URLSessionTask
    func cancel() {
        wrapped.cancel()
    }
}

extension URLSessionTaskWrapper: HTTPClientTask { }
extension URLSessionTaskWrapper: AuthHTTPClientTask { }

final class URLSessionHTTPClient {
    
    private let session: URLSession
    private struct RequestError: Error { }

    init(session: URLSession = .shared) {
        self.session = session
    }
    
    private func dispatch(_ request: URLRequest, _ completion: @escaping (Result<(response: HTTPURLResponse, data: Data), Error>) -> Void) -> URLSessionTaskWrapper {
        let task = session.dataTask(with: request) { data, response, error in
            completion(Result {
                if let error = error {
                    throw error
                } else if let data = data, let response = response as? HTTPURLResponse {
                    return (response, data)
                } else {
                    throw RequestError()
                }
            })
        }
        task.resume()
        return URLSessionTaskWrapper(wrapped: task)
    }
}


extension URLSessionHTTPClient: HTTPClient {
    func execute(_ request: URLRequest, _ completion: @escaping (Result<(response: HTTPURLResponse, data: Data), Error>) -> Void) -> HTTPClientTask {
        return dispatch(request, completion)
    }
}

extension URLSessionHTTPClient: AuthHTTPClient {
    func execute(_ request: URLRequest, _ completion: @escaping (Result<(response: HTTPURLResponse, data: Data), Error>) -> Void) -> AuthHTTPClientTask {
        return dispatch(request, completion)
    }
}

let requestURL = URL(string: "https://jsonplaceholder.typicode.com/todos/1")!

let httpClient: HTTPClient = URLSessionHTTPClient()
httpClient.execute(.init(url: requestURL)) { result in
    if let code = try? result.get().response.statusCode {
        print("HTTP", code)
    }
}

let authzHTTPClient: AuthHTTPClient = URLSessionHTTPClient()
authzHTTPClient.execute(.init(url: requestURL)) { result in
    if let code = try? result.get().response.statusCode {
        print("Authz", code)
    }
}

我如何 return 实现另一个协议的通用类型

How can I return a generic type that implements another protocol

ios

swift

swift-protocols

urlsession