kotlin - 如何在 kotlin` 中使用带有泛型的密封类

Question

我有以下类(class)，但我正在为泛型而苦苦挣扎

sealed class Result<T,E> {
    data class Success<T>(val data: T): Result<T,Void>()
    data class Failure<E>(val error: E): Result<Void,E>()
}


fun interface SeqListener<T,E> {
    fun onComplete(result: Result<T,E>)
}

abstract class Seq<T,E> {

    var started = false
        private set

    var complete = false
        private set

    private lateinit var seqListener: SeqListener<T,E>

    fun start(listener: SeqListener<T,E>) {
        if (!started) {
            seqListener = listener
            started = true
            return doStart()
        } else throw IllegalStateException("cannot call start on a sequence more than once")
    }

    fun success(result: Result.Success<T>) {
        complete(result)
    }

    fun fail(result: Result.Failure<E>) {
        complete(result)
    }

    private fun complete(result: Result<T,E>) {
        if (!complete) {
            complete = true
            seqListener.onComplete(result)
        } else throw IllegalStateException("cannot call success or fail multiple times")
    }

    abstract fun doStart()
}

但编译器提示调用:complete(result)上面的失败/成功属于错误类型 Required: Result<T,E> Found: Result.Success<T>对于成功者。

如果我更改 complete方法:complete(result: Result<*,*>)然后这些错误消失了，但是调用了seqListener.onComplete(result)失败 Required: Result<T,E> Found: Result<*,*>

如有任何帮助，我们将不胜感激。

Answer 1

将您的类型定义为out 并为未使用的类型使用Nothing。这将使编译器对可接受的向上转换非常宽松。

Nothing 类型可以被认为是Any 的对立面，因为它在逻辑上被视为所有内容的子类型。所以对于协变的 out 类型，它可以隐式地向上转换为任何东西。

sealed class Result<out T, out E> {
    data class Success<T>(val data: T): Result<T, Nothing>()
    data class Failure<E>(val error: E): Result<Nothing, E>()
}