generics - 从泛型函数返回特定实例

Question

我的问题是关于泛型方法的类型推断。

我有以下场景:

interface Obj {
    val Id: String
}
data class User(override val Id: String, val name: String): Obj
data class Case(override val Id: String, val subject: String): Obj

interface Api {
    fun <T: Obj> getLastUpdated(type: KClass<T>, backTill: Duration = Duration.ofDays(1)): LastUpdated

    fun <T: Obj> getDetails(type: KClass<T>, uuid: String): Details<T>

    data class LastUpdatedResponse(val ids: List<String> = emptyList(),
                                   val latestDateCovered: String = "")
    data class LastUpdated(val error: Throwable? = null, val response: LastUpdatedResponse? = null)

    data class DetailsResponse<T>(val wrapped: T)
    data class Details<T>(val error: Throwable? = null, val response: DetailsResponse<T>? = null)
}

在我的测试中，我需要确切地知道模拟的 API 需要返回什么，因此

val testCase = Case("123", "Testing")
val testUser = User("321", "Dummy")
val mockApi = object: Api {
    override fun <T : Obj> getLastUpdated(type: KClass<T>, backTill: Duration): Api.LastUpdated {
        return Api.LastUpdated(response = Api.LastUpdatedResponse(listOf("123")))
    }

    override fun <T : Obj> getDetails(type: KClass<T>, uuid: String): Details<T> {
        when (type) {
            Case::class -> return Api.Details(response = Api.DetailsResponse(wrapped = testCase)) // <- this fails
            User::class -> return Api.Details(response = Api.DetailsResponse(wrapped = testUser)) // <- as does this
            else -> return Api.Details(error = UnsupportedOperationException())
        }
    }
}

但是，这无法编译:

Error:(114, 43) Kotlin: Type inference failed. Expected type mismatch: inferred type is Api.Details<Case> but Api.Details<T> was expected

我可以通过强制转换使其工作:

when (type) {
    Case::class -> return Api.Details(response = Api.DetailsResponse(wrapped = testCase)) as Api.Details<T>
    User::class -> return Api.Details(response = Api.DetailsResponse(wrapped = testUser)) as Api.Details<T>

但随后我收到一条警告，通知我“此转换永远不会成功” - 运行我的测试，它确实按预期工作。

我的问题是 - 为什么在没有强制转换的情况下这不起作用？我应该使用什么来代替？

Answer 1

方差注释可能会帮助您，只需将以下函数中的泛型参数 T 更改为 out Obj(在接口(interface)和实现类中):

fun <T: Obj> getDetails(type: KClass<T>, uuid: String): Details<out Obj>