Question

我有一个案例，我想在服务中注册一个参数或没有参数闭包。总有一个参数可用，但为了简洁起见，我希望能够注册没有arg闭包，然后在这种情况下只调度闭包而没有可用的参数。来自强大的OO和动态类型背景，我们喜欢多态分派和类继承树，并让类型自我解决，我可以将以下内容放在一起：

class AbstractAction<T> {
    func publish(value:T) {
        fatalError("you should override this")
    }
}

class NullaryAction<T>: AbstractAction<T> {
    var closure:() -> ()
    override func publish(_:T) {
        closure()
    }
    init(closure:()->()) {
        self.closure = closure
    }
}

class UnaryAction<T>: AbstractAction<T> {
    var closure:(T) -> ()
    override func publish(value:T) {
        closure(value)
    }
    init(closure:(T)->()) {
        self.closure = closure
    }
}

var action:AbstractAction = UnaryAction<Int>(closure: { print("\($0)") })
action.publish(42)
action = NullaryAction<Int>(closure: { print("something happened") } )
action.publish(42)

所以我在控制台中看到42后跟something happened。大。

但我想通过struct和/或enum来探索这一点。价值语义风靡一时。我认为enum方法相对简单：

enum Action<T> {
    case Nullary( ()->() )
    case Unary( (T)->() )

    func publish(value:T) {
        switch self {
        case .Nullary(let closure):
            closure()
        case .Unary(let closure):
            closure(value)
        }
    }
}

var action = Action.Unary({ (arg:Int) -> () in print("\(arg)") })
action.publish(42)
action = Action<Int>.Unary( { print("shorthand too \($0)") } )
action.publish(42)
action = Action<Int>.Nullary({ print("something happened") })
action.publish(42)

要做struct方法，我理解我应该使用协议来捕获publish(value:T)的公共接口。但事情变得令人困惑，因为协议显然不能与泛型相混合？我试过了：

struct NullaryAction<T> {
    typealias ValueType = T
    var closure:() -> ()
}

struct UnaryAction<T> {
    typealias ValueType = T
    var closure:(T) -> ()
}

protocol Action {
    typealias ValueType
    func publish(value:ValueType)
}

extension NullaryAction: Action {
    func publish(_:ValueType) {
        self.closure()
    }
}

extension UnaryAction: Action {
    func publish(value:ValueType) {
        self.closure(value)
    }
}

var action:Action = UnaryAction(closure: { (arg:Int) -> () in print("\(arg)") })
action.publish(42)
action = UnaryAction<Int>(closure: { print("shorthand too \($0)") } )
action.publish(42)
action = NullaryAction<Int>(closure:{ print("something happened") })
action.publish(42)

这只会在底部产生很多错误。我曾尝试将扩展名作为泛型（例如extension NullaryAction<T>:Action），但它告诉我T未被使用，即使我已将typealias表达式放在扩展名中。

是否可以使用struct / protocol执行此操作？我对enum解决方案感到满意，但很失望，我无法用结构/协议方法实现它。

Answer 1

根据您希望将结构转换为其协议（使用var action: Action = UnaryAction {...}）这一事实来判断，我假设您不需要publish方法来获取正确的签名它

换句话说，通过使用Action声明typealias协议，编译器期望为每个结构实例专门化publish方法。

这意味着您有两个选择：

删除类型转换：

示例：

var action /*removed the :Action type casting */ = UnaryAction<Int>(closure: { (arg:Int) -> () in print("\(arg)") })
action.publish(42)
action = UnaryAction<Int>(closure: { print("shorthand too \($0)") } )
action.publish(42)
var anotherAction = NullaryAction<Int>(closure:{ print("something happened") }) //need to use another variable for this last one
anotherAction.publish(42)

此解决方案使您的publish方法也具有与结构相同的签名。如果您的结构专门用于Ints，那么您将拥有.publish(value: Int)。

使协议非通用

示例：

protocol Action {
    func publish(value:Any)
}

struct NullaryAction<T>: Action {
    let closure: () -> ()
    init(closure: () -> ()) {
        self.closure = closure
    }
    func publish(value:Any) {
        self.closure()
    }
}

struct UnaryAction<T>: Action {
    let closure: (T) -> ()
    init(closure: (T) -> ()) {
        self.closure = closure
    }
    func publish(value:Any) {
        self.closure(value as! T) //Need to type cast here
    }
}

var action: Action = UnaryAction<Int>(closure: { (arg:Int) -> () in print("\(arg)") })
action.publish(42)
action = UnaryAction<Int>(closure: { print("shorthand too \($0)") } )
action.publish(42)
action = NullaryAction<Int>(closure:{ print("something happened") })
action.publish(42)

此解决方案允许您继续进行类型转换，但您的publish方法将具有相同的签名（.publish(value: Any)）。您还需要在执行关闭时考虑到这一点。

使用struct / protocols进行Swift多态闭包调度

1 个答案: