问题描述
我有一个通用的REST请求:
I have a generic REST request:
struct Request<T> {…}
T是请求的返回类型，例如:
The T is the return type of the request, for example:
struct Animal {…}
let animalRequest = Request<Animal>
let animal: Animal = sendRequest(animalRequest)
现在，我想表示泛型类型必须符合Decodable的要求，以便我可以解码来自服务器的JSON响应:
Now I would like to express that the generic type has to conform to Decodable so that I can decode the JSON response from the server:
struct Request<T> where T: Decodable {…}
struct Animal: Decodable {…}
这很有意义并且可以工作–直到我收到一个没有响应的请求，即Request<Void>.编译器对此不满意:
This makes sense and works – until I arrive at a request that has no response, a Request<Void>. The compiler is not happy about that one:
Type 'Void' does not conform to protocol 'Decodable'
我通过将Decodable一致性添加到Void来解决这个问题的顽强尝试很快被编译器发现:
My mischevious attempt to solve this by adding the Decodable conformance to Void was quickly found out by the compiler:
extension Void: Decodable {…} // Error: Non-nominal type 'Void' cannot be extended
让请求在返回类型上通用是正确的.有没有办法使它与Void返回类型一起使用? (例如，仅在服务器上创建内容但不返回任何内容的请求.)
It feels right to have the request generic over the return type. Is there a way to make it work with Void return types? (For example the requests that just create something on the server and don’t return anything.)
推荐答案
一个简单的解决方法是引入一个自定义的"no-reply"类型，该类型将替换Void:
A simple workaround is to introduce a custom "no-reply" type that would replace Void:
struct NoReply: Decodable {}
无法将Void符合Decodable. Void只是空元组()的类型别名，并且元组此刻无法遵循协议，但最终会遵循.
Conforming Void to Decodable is not possible. Void is just a type alias for an empty tuple, (), and tuples cannot conform to protocols at this moment, but they will, eventually.
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