如何在Swift中创建通用便利初始化器? [英] How to create generic convenience initializer in Swift?

问题描述

我正在处理Swift中的泛型.我已经扩展到NSManagedObject类，并想创建初始化器，该初始化器仅可用于实现我定义的某些协议的类.现在，我有类似下面的内容，但是它不起作用，甚至无法编译.您能帮我使它正常工作吗?

I'm tackling with generics in Swift. I've got extension to NSManagedObject class and wanted to create initializer which is only available for classes which implements some protocol I defined. Now I've got something like below but this is not working and even not compiling. Could you help me make it working?

public extension NSManagedObject {
    public convenience init<Self: Nameable>(context: NSManagedObjectContext)     {
        let entity = NSEntityDescription.entityForName(Self.entityName(), inManagedObjectContext: context)!
        self.init(entity: entity, insertIntoManagedObjectContext: context)
    }
}

public protocol Nameable {
    static func entityName() -> String
}

Xcode说:函数签名中未使用通用参数'Self'."

Xcode says: "Generic parameter 'Self' is not used in function signature".

推荐答案

正如matt所述，您不能定义一个初始化器 仅限于实现协议的类型.或者，你 可以改为定义全局函数:

As matt already explained, you cannot define an initializer which is restricted to types implementing a protocol. Alternatively, you could define a global function instead:

public protocol Nameable {
    static func entityName() -> String
}

func createInstance<T : NSManagedObject>(type : T.Type, context : NSManagedObjectContext) -> T where T: Nameable {
    let entity = NSEntityDescription.entity(forEntityName: T.entityName(), in: context)!
    return T(entity: entity, insertInto: context)
}

然后用作

let obj = createInstance(Entity.self, context)

如果您定义方法，则可以避免使用其他类型参数

You can avoid the additional type parameter if you define the method as

func createInstance<T : NSManagedObject>(context : NSManagedObjectContext) -> T where T: Nameable { ... }

并将其用作

let obj : Entity = createInstance(context)

或

let obj = createInstance(context) as Entity

现在从上下文中推断出类型.

where the type is now inferred from the context.

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