arrays - 数组<编号> : get and set Int values without casting

Question

我正在构建一个 Matrix 类并希望能够存储 Number s 在二维数组中。

var data: Array<Array<Number>> = Array(width, {Array(height, {0})})

这不起作用，因为 Array<Number>和Array<Int>是不变的。我可以通过使用 Array<Array<out Number>> 使其工作，但矩阵是不可变的，我不希望这样......

选角{0 as Int}使编译器错误消失，但这似乎不是一个好主意。我也想做加法之类的事情，我注意到不可能添加 Number s:

var n: Number = 1
n + 1 // does not work

那么我该如何解决这个问题呢？为什么我不能添加两个 Number是吗？

Answer 1

Number 是一个抽象类，没有定义任何加法。由于没有定义添加数字的方法，因此您无法执行 numberInstane + otherNumberInstance。但是，您可以为其创建一个运算符函数:

infix operator fun Number.plus(other: Number) : Number{
    return when (this) {
        is Double -> this + other.toDouble()
        is Int -> this + other.toInt()
        is Long -> this + other.toLong()
        is Float -> this + other.toFloat()
        is Short -> this + other.toShort()
        is Byte ->  this + other.toByte()
        else -> 0
    }
}

请注意，这仅适用于加法。其余函数将遵循相同的模式，但替换运算符(此处为 +)和函数名称(此处为 plus)。

<小时/>

正如 mer msrd0 的评论，上述结果将导致 1 + 1.5 为 2，因为它向下舍入。 Kotlin 支持相互添加数字类型，最终得到了这个有点可怕的解决方案:

infix operator fun Number.plus(other: Number) : Number{

    when {
        this is Double -> {
            return when(other){
                is Double -> this + other
                is Int -> this + other
                is Long -> this + other
                is Float -> this + other
                is Short -> this + other
                is Byte ->  this + other
                else -> 0
            }
        }
        this is Int -> {
            return when(other){
                is Double -> this + other
                is Int -> this + other
                is Long -> this + other
                is Float -> this + other
                is Short -> this + other
                is Byte ->  this + other
                else -> 0
            }
        }
        this is Long -> {
            return when(other){
                is Double -> this + other
                is Int -> this + other
                is Long -> this + other
                is Float -> this + other
                is Short -> this + other
                is Byte ->  this + other
                else -> 0
            }
        }
        this is Float -> {
            return when(other){
                is Double -> this + other
                is Int -> this + other
                is Long -> this + other
                is Float -> this + other
                is Short -> this + other
                is Byte ->  this + other
                else -> 0
            }
        }
        this is Short -> {
            return when(other){
                is Double -> this + other
                is Int -> this + other
                is Long -> this + other
                is Float -> this + other
                is Short -> this + other
                is Byte ->  this + other
                else -> 0
            }
        }
        this is Byte -> {
            return when(other){
                is Double -> this + other
                is Int -> this + other
                is Long -> this + other
                is Float -> this + other
                is Short -> this + other
                is Byte ->  this + other
                else -> 0
            }
        }
        else -> return 0
    }
}

嵌套的when语句有助于自动转换值，这是必要的，因为Number不是特定的已知类。虽然可能有更好的解决方案，但不知 Prop 体类型。扩展函数主要只是基于类型的自动转换，但不能有单个变量，因为它需要定义为一个数字才能接受所有类型，并且因为有两个变量都需要正确的转换根据传递的类型，它最终会有点困惑。