通用类型错误:无法显式地专门化泛型类型

时间:2017-05-19 07:42:01

标签: swift cocoa-touch generics swift3 protocols

我正在尝试在RequestManager中创建一个通用函数,它通过ServiceManager将收到的JSON从服务器转换为指定的类型。这是我的代码:

RequestManager:

typealias ResultResponseManager  = (_ data: AnyObject?, _ error: ErrorPortage?) -> Void
typealias SuccessResponseManager = (_ success: Bool, _ error: ErrorPortage?) -> Void
typealias objectBlock<T:GenericModel> = (_ object: T?, _ error: ErrorPortage?) -> Void

extension RequestManager {

    static func getObject<T: GenericModel>(endpoint: String, completionBlock: objectBlock<T>? = nil){

        RequestHelper(url: "\(getAPIURL())\(endpoint))")
            .performJSONLaunchRequest { (result, error) in

                if let result = result as? NSDictionary,
                    error == nil {
                    let object = T(dic: result)
                    completionBlock?(object, nil)
                }
                else {
                    completionBlock?(nil, error)
                }
        }
    }


}

ServiceManger:

typealias ObjectResult =  (GenericModel?, ErrorPortage?) -> Void
typealias ObjectsResult =  ([GenericModel]?, ErrorPortage?) -> Void

extension ServiceManager {

    static func getUser(_ id: Int? = nil, _ completion: ObjectResult? = nil) {

        guard let userId: Int = id ?? UserManager.shared.userId else {

            return
        }

        RequestManager.getObject<User>(endpoint: "users/\(userId)") { (user, error) in
            if user = user {

                //update userdefault
                if userId == UserManager.shared.userId {
                    UserDefaults.standard.set(result, forKey: "currentUser")
                }
            }
        }
    }
}

RequestManager.getObject<User> ...行上我收到此错误:

  

无法明确专门化泛型类型

那么我在这里想念的是什么?

  

由于luk2302

,问题解决了

更新 任何想法如何改进这个代码可能或使其更干净! 注意:这不是一个问题,只是关于良好的编程习惯

1 个答案:

答案 0 :(得分:3)

比较https://stackoverflow.com/a/35372990/2442804 - 您不允许“手动”指定类型约束。你必须让编译器推断它。通过以下方式做到:

RequestManager.getObject(endpoint: "users/\(userId)") { (user : User?, error) in 
    // ... your normal code