Question

我对这个ArithmeticException感到有点困惑。我在Scala 2.8.0.RC6和RC7上试过这个。

scala> 7.12 to(8, 0.2)
res0: scala.collection.immutable.NumericRange[Double] = NumericRange(7.12, 7.32, 7.52, 7.72, 7.92)

scala> 7.12 to(8, 0.5)
res2: scala.collection.immutable.NumericRange[Double] = NumericRange(7.12, 7.62)

scala> 7.12 to(8, 0.3)
java.lang.ArithmeticException: Non-terminating decimal expansion; no exact representable decimal result.
 at java.math.BigDecimal.divide(BigDecimal.java:1525)
 at java.math.BigDecimal.divide(BigDecimal.java:1558)
 at scala.math.BigDecimal.$div(BigDecimal.scala:228)
 at scala.math.Numeric$BigDecimalAsIfIntegral$class.quot(Numeric.scala:156)
 at scala.math.Numeric$BigDecimalAsIfIntegral$.quot(Numeric.scala:163)
 at scala.math.Numeric$BigDecimalAsIfIntegral$.quot(Numeric.scala:163)
 at scala.math.Integral$IntegralOps.$div$percent(Integral.scala:23)
 at scala.collection.immutable.NumericRange.genericLength(NumericRange.scala:104)
 at scala.collection.immutable.NumericRange.<init>(NumericRange.scala:63)
 at scala.collection.immutable.NumericRange$Inclusive.<init>(NumericRange.scala:209)
 at ...

Answer 1

这个BigDecimal究竟来自哪里？从Range.scala

开始

  // Double works by using a BigDecimal under the hood for precise
  // stepping, but mapping the sequence values back to doubles with
  // .doubleValue.  This constructs the BigDecimals by way of the
  // String constructor (valueOf) instead of the Double one, which
  // is necessary to keep 0.3d at 0.3 as opposed to
  // 0.299999999999999988897769753748434595763683319091796875 or so.
  object Double {
    implicit val bigDecAsIntegral = scala.Numeric.BigDecimalAsIfIntegral
    implicit val doubleAsIntegral = scala.Numeric.DoubleAsIfIntegral
    def toBD(x: Double): BigDecimal = scala.BigDecimal valueOf x

    def apply(start: Double, end: Double, step: Double) =
      BigDecimal(toBD(start), toBD(end), toBD(step)) mapRange (_.doubleValue)

    def inclusive(start: Double, end: Double, step: Double) =
      BigDecimal.inclusive(toBD(start), toBD(end), toBD(step)) mapRange (_.doubleValue)
  }

转到NumericRange.scala：

  // Motivated by the desire for Double ranges with BigDecimal precision,
  // we need some way to map a Range and get another Range.  This can't be
  // done in any fully general way because Ranges are not arbitrary
  // sequences but step-valued, so we have a custom method only we can call
  // which we promise to use responsibly.
  // 
  // The point of it all is that
  //
  //   0.0 to 1.0 by 0.1
  //
  // should result in
  //
  //   NumericRange[Double](0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0)
  //
  // and not 
  // 
  //   NumericRange[Double](0.0, 0.1, 0.2, 0.30000000000000004, 0.4, 0.5, 0.6000000000000001, 0.7000000000000001, 0.8, 0.9)
  //
  // or perhaps more importantly,
  //
  //   (0.1 to 0.3 by 0.1 contains 0.3) == true
  //
  private[immutable] def mapRange[A](fm: T => A)(implicit unum: Integral[A]): NumericRange[A] = {    
    val self = this

    // XXX This may be incomplete.
    new NumericRange[A](fm(start), fm(end), fm(step), isInclusive) {
      def copy(start: A, end: A, step: A): NumericRange[A] =
        if (isInclusive) NumericRange.inclusive(start, end, step)
        else NumericRange(start, end, step)

      private val underlyingRange: NumericRange[T] = self
      override def foreach[U](f: A => U) { underlyingRange foreach (x => f(fm(x))) }
      override def isEmpty = underlyingRange.isEmpty
      override def apply(idx: Int): A = fm(underlyingRange(idx))
      override def containsTyped(el: A) = underlyingRange exists (x => fm(x) == el)
    }
  }

如果toBD使用允许舍入的MathContext，天空是否会下降？两个邪恶中哪一个较小？我会把这个问题推迟到@extempore。

Answer 2

我answered this上的Scala Forum：

scala> import java.math.{ MathContext => MC, RoundingMode => RM }
import java.math.{MathContext=>MC, RoundingMode=>RM}

scala> val mc1 = new MC(6, RM.HALF_DOWN)
mc1: java.math.MathContext = precision=6 roundingMode=HALF_DOWN


scala> val a1 = BigDecimal(1, mc1)
a1: scala.math.BigDecimal = 1

scala> val b1 = BigDecimal(3, mc1)
b1: scala.math.BigDecimal = 3

scala> a1 / b1
res10: scala.math.BigDecimal = 0.333333


scala> val a2 = BigDecimal(1, MC.DECIMAL128)
a2: scala.math.BigDecimal = 1

scala> val b2 = BigDecimal(3, MC.DECIMAL128)
b2: scala.math.BigDecimal = 3

scala> a2 / b2
res11: scala.math.BigDecimal = 0.3333333333333333333333333333333333

但是，我不确定这是你想要的结果：

scala> def BD128(d: Double): BigDecimal = BigDecimal(d, MC.DECIMAL128)
BD128: (d: Double)BigDecimal

scala> BD128(7.12) to(BD128(8), BD128(0.3))
res10: scala.collection.immutable.NumericRange.Inclusive[BigDecimal] = NumericRange(7.12, 7.42, 7.72)

Answer 3

好的，因为这里什么都没发生，我建议在collection.immutable.Range中进行以下更改：

  object Double {
    implicit val bigDecAsIntegral = scala.Numeric.BigDecimalAsIfIntegral
    implicit val doubleAsIntegral = scala.Numeric.DoubleAsIfIntegral
    def toBD(x: Double): BigDecimal = { 
      // Let's round this Double to prevent
      // error caused by /% in NumericRange
      scala.BigDecimal(x, MC.DECIMAL128)
    }

    def apply(start: Double, end: Double, step: Double) =
      BigDecimal(toBD(start), toBD(end), toBD(step)) mapRange (_.doubleValue)

    def inclusive(start: Double, end: Double, step: Double) =
      BigDecimal.inclusive(toBD(start), toBD(end), toBD(step)) mapRange (_.doubleValue)
  }

我们需要一个解决方案，我不在乎是否会丢失一些微小的精度。

使用RichDouble.to操作获取NumericRange

3 个答案: