结果类型通用参数 'T' 无法推断 swift 5

Question

我正在尝试了解我可以在协议中使用多少泛型。我的想法如下：

protocol Network {
    func fetchCodable<T:Codable>(urlRequest:URLRequest, completion:@escaping (Result<T,Error>)->Void)

}

然后我创建一个名为 AppNetwork 的 class 来实现网络协议。

extension AppNetwork:Network{
    func fetchCodable<T>(urlRequest: URLRequest, completion: @escaping (Result<T, Error>) -> Void) where T : Decodable, T : Encodable {

        URLSession.shared.dataTask(with: urlRequest) { (data, response, error) in

            if let error = error {
                completion(.failure(error))
                return
            }

            guard let data = data else{
                completion(.failure(AppNetworkError.dataCorrupted))
                return
            }

            do {
                let response = try JSONDecoder().decode(T.self, from: data)
                completion(.success(response))
            }
            catch let decodeError{
                completion(.failure(decodeError))
            }

        }.resume()


    }

此 class 是 NetworkHelper 的一部分，它实现了检索数据的功能，例如：

class NetworkHelper{
    let network:Network = AppNetwork()
}
...
//MARK:- Public methods
extension NetworkHelper{

    func getVenueDetails(inLocation:String, offset:Int, limit:Int, radius:Int = 1000,completion:(Result<VenueDetailsSearchResult, Error>)->Void){

        guard let foursquareConfig = foursquareConfig else{
            completion(.failure(NetworkHelper.NetworkHelperError.invalidFourSquareConfig))
            return
        }

        var venuesURLString = EndPoints.venueSearch.rawValue + "?"
        venuesURLString += foursquareConfig.getFormattedParams
        venuesURLString += "&near=\(inLocation)&radius=\(radius)&offset=\(offset)&limit=\(limit)"

        guard let venuesURL = URL(string: venuesURLString) else{
            completion(.failure(NetworkHelper.NetworkHelperError.invalidURL))
            return
        }

        let venuesURLRequest = URLRequest(url: venuesURL, cachePolicy: .reloadIgnoringLocalCacheData, timeoutInterval: 10)
        network.fetchCodable(urlRequest: venuesURLRequest) { result in



        }

    }

}

在这种情况下，我试图获得一个结果，但我收到一条消息错误："Generic parameter 'T' could not be inferred"

如果我能做到，我应该怎么做？

谢谢

Answer 1

您必须将您期望的类型告诉编译器。

network.fetchCodable(urlRequest: venuesURLRequest) { (_ result: Result<VenueDetailsSearchResult, Error>) in
    // ...
}

结果类型通用参数 'T' 无法推断 swift 5

Result type Generic parameter 'T' could not be inferred swift 5

generics

swift

codable

swift5