排列 <Number> ：获取并设置Int值而不进行强制转换

Question

我正在构建一个Matrix类，并希望能够将Number存储在二维数组中。

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>>来使它工作，但Matrix将是不可变的，我不希望这样......

将{0 as Int}会使编译器错误消失，但这似乎不是一个好主意。 我也想做像添加这样的事情，我注意到它不可能添加Number s：

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

那么我怎么能解决这个问题呢？ 为什么我不能添加两个Number ？

Answer 1

Number是一个抽象类，不会为添加定义任何内容。 由于没有用于添加数字的已定义方法，因此您无法执行numberInstane + otherNumberInstance 。 但是，您可以为它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，这仅适用于添加。 其余的函数将遵循相同的模式，但替换运算符（这里是+ ）和函数的名称（这里是plus ）。

---

作为mer msrd0的评论，上面会导致1 + 1.5为2，因为它向下舍入。 Kotlin支持在彼此之间添加数字类型，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语句有助于自动更新值，这是必需的，因为Number不是特定的已知类。 可能有更好的解决方案，但不知道具体的类型。 扩展函数大多只是基于类型的自动转换，但是不能有单个变量，因为它需要被定义为一个数字来接受所有类型，并且因为有两个vars都需要适当的转换根据传递的类型，它最终会有点凌乱。