kotlin - List > + List = List <任何>？

Question

我有下面的代码有效。

class ListManipulate(val list: List<Char>, val blockCount: Int) {

    val result: MutableList<List<Char>> = mutableListOf()

    fun permute(sequence: List<Int> = emptyList(), start: Int = 0, count: Int = blockCount) {
        if (count == 0) {
            result.add(constructSequence(sequence))
            return
        }
        for (i in start .. list.size - count) {
            permute(sequence + i, i + 1, count - 1)
        }
    }

    private fun constructSequence(sequence: List<Int>): List<Char> {
        var result = emptyList<Char>()
        for (i in sequence) {
            result += list[i]
        }
        return result
    }
}

但是，当我将 result从MutableList更改为普通列表时，即

    var result: List<List<Char>> = emptyList()
    // ...
            result += constructSequence(sequence)

我收到此错误 Type mismatch. Require: List<List<Char>>; Found: List<Any>
完整代码如下

class ListManipulate(val list: List<Char>, val blockCount: Int) {

    var result: List<List<Char>> = emptyList()

    fun permute(sequence: List<Int> = emptyList(), start: Int = 0, count: Int = blockCount) {
        if (count == 0) {
            result += constructSequence(sequence)
            return
        }
        for (i in start .. list.size - count) {
            permute(sequence + i, i + 1, count - 1)
        }
    }

    private fun constructSequence(sequence: List<Int>): List<Char> {
        var result = emptyList<Char>()
        for (i in sequence) {
            result += list[i]
        }
        return result
    }
}

为什么 result + constructSequence(sequence)会导致 List<Any>而不是 List<List<Char>>？

有没有办法我仍然可以使用普通的List>而不是可变列表？

Answer 1

CTRL +单击IDEA中的+，您将看到它带您进入以下功能:

/**
 * Returns a list containing all elements of the original collection and then all elements of the given [elements] collection.
 */
public operator fun <T> Collection<T>.plus(elements: Iterable<T>): List<T> {
    /* ... */
}

这意味着您将 elements的所有单个元素添加到接收器。也就是说，您将所有 T都添加到 List<List<T>>中。由于 List<T>不是 T，因此将得到 List<Any>。