您好，我想構建如下所示的內容：
protocol BaseEntity: Identifiable {
var id: UUID { get set }
}
protocol DataStore {
associatedtype T: BaseEntity
typealias CompleteHandler<T> = ((Result<T, Error>) -> Void)
func read(with id: T.ID, completion: CompleteHandler<T>)
}
但在具體實作中，我想利用協議UUID中定義的BaseEntity：
class Storage<Entity: BaseEntity>: DataStore {
func read(with id: Entity.ID, completion: CompleteHandler<Entity>) {
// here, how to use that id as a UUID?
}
}
我不知道該怎么做，定義必須有一個物件符合BaseEntity并希望UUID id在具體實作中使用該屬性。我不想這樣做：
func read(with id: Entity.ID, completion: CompleteHandler<Entity>) {
guard let uuid = id as? UUID else {
return
}
// more code goes here...
}
我應該怎么做？
uj5u.com熱心網友回復：
好吧，基本上在這個實作中缺少任何具體型別是編譯器找出Entity.ID型別的麻煩。
這可以通過Entity在您的實作中不使用泛型型別來顯示Storage，如下所示：
struct MyEntity: BaseEntity {
var id: UUID
}
class Storage: DataStore {
func read(with id: MyEntity.ID, completion: CompleteHandler<MyEntity>) {
print(id.uuidString) // <- `id` is of `UUID` type
}
}
解決方案 1：為了幫助編譯器理解Entity.ID型別，您可以有條件地遵守DataStore協議，宣告它Entity.ID始終是UUID型別（這實際上是您在實作中想要的）：
class Storage<Entity: BaseEntity>: DataStore where Entity.ID == UUID {
func read(with id: Entity.ID, completion: CompleteHandler<Entity>) {
print(id.uuidString) // again this line passes compilation
}
}
解決方案 2：更好的是，您可以BaseEntity通過宣告它ID始終是UUID型別來條件約束您的協議。這樣你就不需要改變你的Storage實作：
protocol BaseEntity: Identifiable where ID == UUID {}
// or using the following which is also the same
protocol BaseEntity: Identifiable<UUID> {}
class Storage<Entity: BaseEntity>: DataStore {
func read(with id: Entity.ID, completion: CompleteHandler<Entity>) {
print(id.uuidString) // again this line passes compilation
}
}
解決方案 3：另一種可能更適合您的背景關系的解決方案是轉換BaseEntity為類。這樣你也不需要改變你的Storage實作：
class BaseEntity: Identifiable {
var id: UUID
init(id: UUID) {
self.id = id
}
}
class Storage<Entity: BaseEntity>: DataStore {
func read(with id: Entity.ID, completion: CompleteHandler<Entity>) {
print(id.uuidString) // again this line passes compilation
}
}
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