从 Swift 5 类型<Result>转换为 Swift 3 等效版本

Question

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

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()
}
}

什么相当于 Result.Type in Swift 3func 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

Swift 5 中的Result类型基本上是

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

如果你不需要init(catching或get()功能或map的东西，那么基本的枚举就足够了