我怎样才能在 Swift 中解决 "Generic parameter 'T' could not be inferred"

Question

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
}

我想像那样调用 postAsync func 但它说 **Generic parameter 'TResponse' could not be inferred** 我如何调用这个方法？我尝试了不同的方法，但没有解决。

 - let res = await postAsync(route: "MyController/Get", request: someRequest) as? SomeResponse
 - let res: SomeResponse = await postAsync(route: "MyController/Get", request: someRequest)

Answer 1

我不确定您尝试的解决方案为何失败，也许它们不适用于 async/await 功能。

也就是说，虽然通常可以使用隐式泛型，但 Swift 中更常见的做法是将类型作为参数之一，而不是将其保持隐式。在类中就是这样做的，比如 JSONDecoder。

我建议将您的函数签名更改为：

public func postAsync<TRequest: Codable, TResponse: Codable>(route: String, request: TRequest, receive: TResponse.Type) async throws -> TResponse? {

并这样称呼它：

let res = await postAsync(route: "MyController/Get", request: someRequest, receive: SomeResponse.self)

Answer 2

您的函数不是 return SomeResponse，它 return 是 SomeResponse?，所以您的意思是：

let res = ... as SomeResponse? // Note `as`, not `as?`

或

let res: SomeResponse? = ...

我同意 EmilioPaleaz 关于如何改进 API 的观点，但我建议添加一个默认值，这样可以两全其美：

... request: TRequest, returning: TResponse.Type = TResponse.self) async throws -> ...

有了这个，当return类型已知时，你可以省略它。

我怎样才能在 Swift 中解决 "Generic parameter 'T' could not be inferred"

How can I solve "Generic parameter 'T' could not be inferred" in Swift

generics

swift

swiftui

swift5.5