找到下面的代码
//动物可以吃协议动物{相关类型食物 func feed（食物：食物） - > Void } struct AnyAnimal< Food>：Animal { private let _feed：（Food） - > Void init< Base：Animal where Food == Base.Food>（_ base：Base）{ _feed = base.feed } func feed（food：Food） {_feed（food）} } //各种食物 struct Grass {} struct Cow：Animal { func feed（food：Grass）{print（moo）} } struct Goat：Animal { func feed（food：Grass）{print（bah） } } 让grassEaters = [AnyAnimal（Cow（）），AnyAnimal（Goat（））] for grassEaters { animal.feed（Grass （））}现在我想在协议Animal中给出一个默认实现以下
扩展动物{ func feed（food：Food） - > Void { print（unknown）} }当我从结构牛中删除函数时，我得到了Cow不符合协议动物的错误消息。
这是否意味着您不能使用Type Erasures和默认实现在一起？有什么办法可以做TypeErasure并保持默认实现？
解决方案问题与类型擦除无关，如果您删除 struct AnyAnimal< Food> 定义，则会得到相同的错误消息。
如果您从 struct Cow 然后中删除 feed（）方法，编译器无法推断关联的类型食品。因此，无论您在默认实现中使用了一个具体类型：
extension动物{ func feed食物：草） - > Void { print（unknown）} } struct Cow：Animal {}或者使用的默认实现为每个类型定义一个类型别名 Food ：
extension动物{ func feed（food：Food） - > Void { print（unknown）} } struct Cow：Animal { typealias Food = Grass }也可以为 Food 在协议中：
协议动物{ associatedtype Food = Grass func feed （食物：食物） - >无效} 扩展动物{ func feed（food：Food） - > Void { print（unknown）} } struct Cow：Animal {}
I have been struggling very hard with Swift Protocols and Associated Types for a long time. I started again with basic to really understand what is going wrong and I followed this article of TypeErasure in Swift Protocols with Associated Type Requirement by Rob Napier but still i have no luck.
Find the code below
// An Animal can eat protocol Animal { associatedtype Food func feed(food: Food) -> Void } struct AnyAnimal<Food>: Animal { private let _feed: (Food) -> Void init<Base: Animal where Food == Base.Food>(_ base: Base) { _feed = base.feed } func feed(food: Food) { _feed(food) } } // Kinds of Food struct Grass {} struct Cow: Animal { func feed(food: Grass) { print("moo") } } struct Goat: Animal { func feed(food: Grass) { print("bah") } } let grassEaters = [AnyAnimal(Cow()), AnyAnimal(Goat())] for animal in grassEaters { animal.feed(Grass()) }Now i wanted to give a default implementation in the protocol Animal like below
extension Animal { func feed(food: Food) -> Void { print("unknown") } }And when i removed the function from the struct Cow i get the error message that Cow doesn't conform to Protocol Animal.
Does that mean you cannot have Type Erasures and Default Implementation together?. Is there any way where i can do TypeErasure and also keep a Default Implementation?
解决方案The problem is unrelated to the type erasure and you'll get the same error message if you remove the struct AnyAnimal<Food> definition.
If you remove the feed() method from struct Cow then the compiler cannot infer the associated type Food. So either you use a concrete type in the default implementation:
extension Animal { func feed(food: Grass) -> Void { print("unknown") } } struct Cow: Animal { }or you define a type alias Food for each type using the default implementation:
extension Animal { func feed(food: Food) -> Void { print("unknown") } } struct Cow: Animal { typealias Food = Grass }It is also possible to define a default type for Food in the protocol:
protocol Animal { associatedtype Food = Grass func feed(food: Food) -> Void } extension Animal { func feed(food: Food) -> Void { print("unknown") } } struct Cow: Animal { }
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