问题描述
我有以下结构和函数.
struct ApiResponse<TResponse: Codable>: Codable {
var isSuccess: Bool
var mainResponse: TResponse?
}
public struct SomeResponse: Codable {
var someProperty: String
}
public func postAsync<TRequest: Codable, TResponse: Codable>(route: String, request: TRequest) async throws -> TResponse? {
let body = try JsonEncoder().encode(request)
let urlRequest = createUrlRequest(route: route, method: "POST", body: body)
let (data, _) = try await URLSession.shared.data(for: urlRequest)
let apiResponse = try JsonDecoder().decode(ApiResponse<TResponse>.self, from: data)
return response.mainResponse
}
我想像这样调用 postAsync
func 但它说 **Generic parameter 'TResponse' could not be inferred**
我怎样才能调用这个方法?我尝试了不同的方法但没有解决.
I want to call postAsync
func like that but it says **Generic parameter 'TResponse' could not be inferred**
How can I call this method? I tried different ways but not solve.
- let res = await postAsync(route: "MyController/Get", request: someRequest) as? SomeResponse
- let res: SomeResponse = await postAsync(route: "MyController/Get", request: someRequest)
推荐答案
你的函数不返回 SomeResponse
,它返回 SomeResponse?
,所以你这里的意思是:
Your function doesn't return SomeResponse
, it returns SomeResponse?
, so what you meant here is:
let res = ... as SomeResponse? // Note `as`, not `as?`
或
let res: SomeResponse? = ...
我同意 EmilioPaleaz 关于如何改进 API 的意见,但我建议添加一个默认值,这样可以两全其美:
I agree with EmilioPaleaz about how to improve the API, though I would recommend adding a default value, which gives the best of both worlds:
... request: TRequest, returning: TResponse.Type = TResponse.self) async throws -> ...
有了这个,当返回类型已知时,你可以省略它.
With this, when the return type is known, you can omit it.