为什么Kotlin没有十进制进展？

Question

最近我遇到了一个问题,通过十进制步骤迭代十进制数字,我很想知道,为什么Kotlin只有进步Int,Long而且Char.

我明白,可能会有一些带小数的警告.但仍然.我们只想要一个起始BigDecimal编号,结束BigDecimal编号,然后用BigDecimal步骤迭代它们.

问:那么,为什么整数不存在任何进展？谢谢.

PS:这是一个可能实现的示例代码(我为Int采用了源代码并进行了调整BigDecimal):

/**
 * Returns a progression that goes over the same range with the given step.
 */
public infix fun BigDecimalProgression.step(step: BigDecimal): BigDecimalProgression {
    if (step <= java.math.BigDecimal.ZERO) throw IllegalArgumentException("Step must be positive, was: $step.")
    return BigDecimalProgression.fromClosedRange(first, last, if (this.step > java.math.BigDecimal.ZERO) step else -step)
}

/**
 * A progression of values of type `BigDecimal`.
 */
public open class BigDecimalProgression
internal constructor
(
        start: BigDecimal,
        endInclusive: BigDecimal,
        step: BigDecimal
) : Iterable<BigDecimal> {
    init {
        if (step == BigDecimal.ZERO) throw kotlin.IllegalArgumentException("Step must be non-zero")
    }

    /**
     * The first element in the progression.
     */
    public val first: BigDecimal = start

    /**
     * The last element in the progression.
     */
    public val last: BigDecimal = getProgressionLastElement(start, endInclusive, step)

    /**
     * The step of the progression.
     */
    public val step: BigDecimal = step

    override fun iterator(): BigDecimalIterator = BigDecimalProgressionIterator(first, last, step)

    /** Checks if the progression is empty. */
    public open fun isEmpty(): Boolean = if (step > BigDecimal.ZERO) first > last else first < last

    override fun equals(other: Any?): Boolean =
            other is BigDecimalProgression && (isEmpty() && other.isEmpty() ||
                    first == other.first && last == other.last && step == other.step)

    override fun hashCode(): Int =
            if (isEmpty()) -1 else (31 * (31 * first.hashCode() + last.hashCode()) + step.hashCode())

    override fun toString(): String = if (step > BigDecimal.ZERO) "$first..$last step $step" else "$first downTo $last step ${-step}"

    companion object {
        /**
         * Creates BigDecimalProgression within the specified bounds of a closed range.

         * The progression starts with the [rangeStart] value and goes toward the [rangeEnd] value not excluding it, with the specified [step].
         * In order to go backwards the [step] must be negative.
         */
        public fun fromClosedRange(rangeStart: BigDecimal, rangeEnd: BigDecimal, step: BigDecimal): BigDecimalProgression = BigDecimalProgression(rangeStart, rangeEnd, step)
    }
}

fun getProgressionLastElement(start: BigDecimal, end: BigDecimal, step: BigDecimal): BigDecimal {
    if (step > BigDecimal.ZERO) {
        return start + BigDecimal(((end - start) / step).toInt()) * step
    } else if (step < BigDecimal.ZERO) {
        return start - BigDecimal(((start - end) / -step).toInt()) * -step
    } else {
        throw kotlin.IllegalArgumentException("Step is zero.")
    }
}

/** An iterator over a sequence of values of type `BigDecimal`. */
public abstract class BigDecimalIterator : Iterator<BigDecimal> {
    override final fun next() = nextBigDecimal()

    /** Returns the next value in the sequence without boxing. */
    public abstract fun nextBigDecimal(): BigDecimal
}

/**
 * An iterator over a progression of values of type `BigDecimal`.
 * @property step the number by which the value is incremented on each step.
 */
internal class BigDecimalProgressionIterator(first: BigDecimal, last: BigDecimal, val step: BigDecimal) : BigDecimalIterator() {
    private val finalElement = last
    private var hasNext: Boolean = if (step > BigDecimal.ZERO) first <= last else first >= last
    private var next = if (hasNext) first else finalElement

    override fun hasNext(): Boolean = hasNext

    override fun nextBigDecimal(): BigDecimal {
        val value = next
        if (value >= finalElement) {
            if (!hasNext) throw kotlin.NoSuchElementException()
            hasNext = false
        }
        else {
            next += step
        }
        return value
    }
}

Answer 1

正如范围文档中所述：

浮点数（Double、Float）没有定义其 rangeTo 运算符，而是使用标准库为泛型 Comparable 类型提供的运算符：

public operator fun <T: Comparable<T>> T.rangeTo(that: T): ClosedRange<T>

该函数返回的范围不能用于迭代。由于无法使用范围，因此您将不得不使用其他类型的循环。

他们根本就没有定义。