从 Swift 5 类型 <Result> 转换为 Swift 3 等价

Question

我正在尝试将一些 Swift 5 转换为 Swift 3，因为我遇到向后兼容性问题。

import Foundation
class WS {
    enum WebSessionError : Error {
        case badResponse(String)
    }
    // RequestURL -> API Location
    static let requestURL = URL(string:"-Placeholder-")!
    static var sharedInstance = WS() // Instancing our class
    func run(completion : @escaping (Result<String,Error>) -> Void) {
        let instancedTask = URLSession.shared.dataTask(with: WS.requestURL)
        { (data,response,error) in
            if let error = error {
                print("Client Error: \(error.localizedDescription)")
                completion(.failure(error))
                return
            }
            guard let response = response as? HTTPURLResponse, (200...299).contains(response.statusCode)
                else {
                    completion(.failure(WebSessionError.badResponse("Server Error!")))
                return
            }
            guard let mime = response.mimeType, mime == "text/html" else {
                    completion(.failure(WebSessionError.badResponse("Wrong mime type!")))
                return
            }
            completion(.success(String(data: data!, encoding: .utf8)!))
        }
        instancedTask.resume()
    }
}

Swift 中的 Result.Type 相当于什么 3 func run(completion : @escaping (Result<String,Error>) -> Void) 这部分是我收到构建错误的地方，转换其余部分基本上没问题。

Answer 1

Result类型Swift5基本就是

enum Result<Success, Failure> where Failure : Error {
    case success(Success), failure(Failure)
}

如果您不需要 init(catching 或 get() 功能或 map 基本枚举就足够了

从 Swift 5 类型 <Result> 转换为 Swift 3 等价

Converting From Swift 5 Type <Result> to Swift 3 Equivalent

swift

swift3

swift5