如何使用类型约束调用协议扩展默认实现

Question

考虑以下示例：

class ManObj {
    func baseFunc() {
        print("ManObj baseFunc")
    }
}

class SubObj: ManObj {
}

protocol Model {
}

extension Model { // This is protocol extension
    func someFunc() { // Protocol extension default implementation
        (self as! ManObj).baseFunc()
        print("Model implementation")
    }
}
extension SubObj: Model {
    func someFunc() {
        print("SubObj Implementation")
    }
}

let list = SubObj()
list.someFunc() // static dispatching

let list2: Model = SubObj()
list2.someFunc() // dynamic dispatching

输出很好：

SubObj Implementation
ManObj baseFunc
Model implementation

但我不喜欢 (self as! ManObj).baseFunc().

行中的转换

事实上，我只打算将Model协议应用于ManObj的子类。（但并非 ManObj 的所有子类都是 Model！）因此，我尝试将 Model 更改为：

extension Model where Self: ManObj {
    func someFunc() {
        self.baseFunc() // No more casting needed!
        print("Model implementation")
    }
}

但是我遇到了错误：

list2.someFunc() <- error: 'Model' is not a subtype of 'ManObj'

那么，在我将 Model 限制为 where Self: ManObj 之后，有没有办法让我从 list2 触发 Model.someFunc？

Answer 1

为类型转换创建一个空的class

class ManObj {
    func baseFunc() {
        print("ManObj baseFunc")
    }
}

class SubObj: ModelCaster {
    func someFunc() {
        print("SubObj Implementation")
    }
}

protocol Model {
}

extension Model where Self: ModelCaster { // This is protocol extension
    func someFunc() { // Protocol extension default implementation
        print("Model implementation")
    }
}

class ModelCaster: ManObj, Model{

}

let list = SubObj()
list.someFunc() //SubObj Implementation

let list2: ModelCaster = SubObj()
list2.someFunc() //Model implementation

如何使用类型约束调用协议扩展默认实现

How to invoke protocol extension default implementation with type constraints

swift

protocol-extension