Swift: 如何让泛型代码更干净

Question

我正在寻找使我的代码更简洁、更高效的方法。我有一个“BaseInfo”，它有一些属性，而且它是“SomeInfo”和“AnotherInfo”classes 的父级，它们有自己的附加属性。我创建了一个通用函数，用于从 UserDefaults 中保存和获取这些对象。考虑到它应该使用的信息类型，我有 ViewController 使用这些函数保存和加载信息。我想知道有什么方法可以使我的代码更清晰并摆脱我的 ViewContoller 中的那些类型转换。这是我的信息 classes:

public class  BaseInfo: Codable{
    var name: String?
    
    init(_ dict: [String: Any]){
        
        name = dict["name"] as? String
    }
}

class AnotherInfo: BaseInfo {
    var secondName: String?
    
    override init(_ dict: [String : Any]) {
        super.init(dict)
        secondName = dict["secondName"] as? String
    }
    
    required init(from decoder: Decoder) throws {
        fatalError("init(from:) has not been implemented")
    }
}

class SomeInfo: BaseInfo {
    var middleName: String?
    
    
    override init(_ dict: [String : Any]) {
        super.init(dict)
        middleName = dict["middleName"]
    }
    
    required init(from decoder: Decoder) throws {
        fatalError("init(from:) has not been implemented")
    }
}

这是我的 class 管理信息

protocol InfoManagerProtocol {
    static func getInfo(with type: InfoType) -> BaseInfo?
    static func saveInfo <T: BaseInfo>(with type: InfoType, info: T)
}

class InfoManager: InfoManagerProtocol {
    
    static func getInfo<T: BaseInfo>(with type: InfoType) -> T? {
        if let data = UserDefaults.standard.object(forKey: type.toString()) as? Data, let info = try? JSONDecoder().decode(type.typeOfInfo() as! T.Type, from: data){
            return info
        }
        return nil
    }
    
    
    static func saveInfo<T>(with type: InfoType, info: T) where T : BaseInfo {
        if let encodedData = try? JSONEncoder().encode(info){
            UserDefaults.standard.setValue(encodedData, forKey: type.toString())
        }
    }
}

InfoType 枚举：

enum InfoType: Int{
    case base = 1
    case another = 2
    case some = 3
}

extension InfoType{

    func toString() -> String{
      switch self{
       case .base: 
           return "base"
       case .another: 
           return "another"
       case .some:
           return "some"
      }
    } 
    func typeOfInfo() -> BaseInfo.Type{
        switch self{
        case .base:
            return BaseInfo.self
        case .another:
            return AnotherInfo.self
        case .some:
            return SomeInfo.self
        }
    }
}

和一些控制器

class ViewCV: UIViewController{
    ......
    // some outlets
    var infoType: InfoType? // info 

    // some func that updates UI
    func updateUI(){
       
       let genericInfo = InfoManager.getInfo(info: .infoType)
       self.someNameOutletLabel.text = genericInfo.name
       self.secondNameOutletLabel?.text = (genericInfo as? AnotherInfo).secondName // here I don't like it
       if let someInfo = genericInfo as? SomeInfo{
          self.secondNameOutletLabel?.text = someInfo.thirdName // or like here I also don't like
       }
   }
}

期待其他批评家和建议

Answer 1

您可以通过跳过 InfoType 类型来简化您的代码，而是定义从给定参数或 return 值使用的类型。

所以协议变成

protocol InfoManagerProtocol {
    static func getInfo<T: BaseInfo>() -> T?
    static func saveInfo <T: BaseInfo>(info: T)
}

然后执行。请注意，我现在使用类名本身作为键

而不是枚举中的 toString

class InfoManager: InfoManagerProtocol {
    static func getInfo<T: BaseInfo>() -> T? {
        if let data = UserDefaults.standard.object(forKey: "\(T.self)") as? Data, let info = try? JSONDecoder().decode(T.self, from: data){
            return info
        }
        return nil
    }

    static func saveInfo<T>(info: T) where T : BaseInfo {
        if let encodedData = try? JSONEncoder().encode(info){
            print(encodedData)
            UserDefaults.standard.set(encodedData, forKey: "\(T.self)")
        }
    }
}

下面是如何使用它的例子（假设init(from:)已经正确实施）

let dict: [String: String] = ["name": "Joe", "secondName": "Doe", "middleName": "Jr"]
let another = AnotherInfo(dict)
let some = SomeInfo(dict)

//Here the declaration of the argument tells the generic function what T is
InfoManager.saveInfo(info: another)
InfoManager.saveInfo(info: some)

//And here the declaration of the return value tells the generic function what T is
if let stored:AnotherInfo = InfoManager.getInfo() {
    print(stored, type(of: stored))
}
if let stored:SomeInfo = InfoManager.getInfo() {
    print(stored, type(of: stored))
}

Swift: 如何让泛型代码更干净

Swift: How to make Generics code more clean

ios

swift

generics

types

type-alias