numbers - Kotlin:有没有办法清除数字重载？

Question

我经常发现自己为数字类型创建了很多重载，因为Kotlin不会对非文字进行隐式转换。这导致大量重复的过载功能，即成千上万的样板。

我在此遇到的痛苦的一个例子可以在这里看到:https://github.com/Jire/Arrowhead/blob/master/src/main/kotlin/org/jire/arrowhead/Source.kt

我理解隐式转换会导致bug的原因，但是我认为在大多数情况下，尤其是对于不丢失数据的Byte -> Int和Int -> Long之类的“扩大”转换，应该有更好的方法。

所以...我的问题是您如何处理？有什么创造性的方法可以解决这个问题？

Answer 1

回答您的问题:可以将通用函数与 Number 一起使用。然后它将接受Number的任何子类型。然后，您可以将值转换为任何其他数字类型:

fun boo(x: Number) {
    val y = x.toLong()
    println(y is Long)
} 

唯一的缺点是自动装箱，但这与您的情况无关。

关于您发布的代码:我相信您有一些架构错误会导致这种情况。这是您的API，没有太多详细信息:

interface Source {
    fun read(address: Long, data: Pointer, bytesToRead: Int) 

    fun read(address: Int, data: Pointer, bytesToRead: Int) = read(address.toLong(), data, bytesToRead)
    fun read(address: Long, data: Memory, bytesToRead: Int = data.size().toInt()) = read(address, data as Pointer, bytesToRead)
    fun read(address: Int, data: Memory, bytesToRead: Int = data.size().toInt()) = read(address.toLong(), data, bytesToRead)
    fun read(address: Long, struct: Struct, bytesToRead: Int = struct.size()) = read(address, struct.pointer, bytesToRead)
    fun read(address: Int, struct: Struct, bytesToRead: Int = struct.size()) = read(address.toLong(), struct, bytesToRead)
    fun read(address: Long, bytesToRead: Int): Memory = TODO()
    fun read(address: Int, bytesToRead: Int) = read(address.toLong(), bytesToRead)


    fun byte(address: Long, offset: Long = 0) = read(address, 1).getByte(offset)
    fun byte(address: Int, offset: Long = 0) = byte(address.toLong(), offset)

    fun short(address: Long, offset: Long = 0) = read(address, 2).getShort(offset)
    fun short(address: Int, offset: Long = 0) = short(address.toLong(), offset)

    fun char(address: Long, offset: Long = 0) = read(address, 2).getChar(offset)
    fun char(address: Int, offset: Long = 0) = char(address.toLong(), offset)

    fun int(address: Long, offset: Long = 0) = read(address, 4).getInt(offset)
    fun int(address: Int, offset: Long = 0) = int(address.toLong(), offset)

    fun long(address: Long, offset: Long = 0) = read(address, 8).getLong(offset)
    fun long(address: Int, offset: Long = 0) = long(address.toLong(), offset)

    fun float(address: Long, offset: Long = 0) = read(address, 4).getFloat(offset)
    fun float(address: Int, offset: Long = 0) = float(address.toLong(), offset)

    fun double(address: Long, offset: Long = 0) = read(address, 8).getDouble(offset)
    fun double(address: Int, offset: Long = 0) = double(address.toLong(), offset)

    fun boolean(address: Long, offset: Long = 0) = byte(address, offset).unsign() > 0
    fun boolean(address: Int, offset: Long = 0) = boolean(address.toLong(), offset)

    fun write(address: Long, data: Pointer, bytesToWrite: Int)
    fun write(address: Int, data: Pointer, bytesToWrite: Int) = write(address.toLong(), data, bytesToWrite)
    fun write(address: Long, data: Memory, bytesToWrite: Int = data.size().toInt())
            = write(address, data as Pointer, bytesToWrite)
    fun write(address: Int, data: Memory, bytesToWrite: Int = data.size().toInt())
            = write(address.toLong(), data, bytesToWrite)
    fun write(address: Long, struct: Struct, bytesToWrite: Int = struct.size())
            = write(address, struct.pointer, bytesToWrite)
    fun write(address: Int, struct: Struct, bytesToWrite: Int = struct.size())
            = write(address.toLong(), struct, bytesToWrite)

    // ...
}  

此API与Long一起用于address，但由于某种原因也接受Int。我认为您应该偷看一个(例如Long)，让消费者担心将Int转换为Long。这不是API的责任。此外，如果消费者使用将Long用作address的API，那么他通常也将使用Long来处理自己身边的地址。这简化了API，并节省了Int到Long以及向后转换的时间，从而提高了性能。