如何使用 Swift 中的调用者类型推断静态泛型方法中的类型

Question

我在协议扩展上声明此方法

protocol Storable { ... }

extention Storable {
    static func get<T: Decodable>(by identifier: String, completion: @escaping (T?) -> Void)
    ...
}

现在我在实现 Storable.

的类型上使用该方法

struct User: Storable {
    ...
}

User.get(by: "userId", completion: { user in
    print(user)
}

但是编译器说：Generic parameter 'T' could not be inferred

我想告诉编译器“T是调用静态方法的class”

我编译成功:

static func get<T>(by identifier: String, type: T.Type, completion: @escaping (T?) -> Void) where T: Decodable

and 

User.get(by: "mprot", type: User.self) { ... }

不过好像有点多余:(

Answer 1

我想告诉编译器"T is the class who calls the static method"

假设您只想在 T 是调用静态方法的 class 时应用您的 get，这是怎么回事？

protocol Storable { //... } extension Storable where Self: Decodable { static func get(by identifier: String, completion: @escaping (Self?) -> Void) { //... } } struct User: Storable, Decodable { //... }

这将编译成功：

User.get(by: "userId", completion: { user in print(user) })

Answer 2

T 不能是可选的。你应该有：

static func get<T: Decodable>(by identifier: String, completion: @escaping (T) -> ())

被调用

User.get(by: "userId", completion: { (value: User) in ... })

如何使用 Swift 中的调用者类型推断静态泛型方法中的类型

How to infer type in static generic method with the type of the caller in Swift

generics

static

protocols

swift

我想告诉编译器"T is the class who calls the static method"