scala - Scala Numeric[T] 的问题

Question

我有一个数字求解函数 (Double => Double)，我试图聪明地使用 Numeric[T] 来将两种数字分开.

事实证明这并不容易。剩下的问题是:

• 如何做除法； Numeric[T] 只有加号、减号等运算符。
• 为什么编译器找不到隐含证据$1: Numeric[Double] 函数(参见下面的编译器输出)

理想情况下，我想说，“A 和 B 都是 Double，但如果我将它们混合在一起，请告诉我其他”。

代码如下:

import scala.annotation.tailrec

class Sweep[A: Numeric, B: Numeric]( fDiff: A => B, initialSeed: A, initialStep: A, bEps: B )
{
  val anum= evidence$1
  val bnum= evidence$2

  assert( anum.signum(initialStep) > 0 )
  assert( bnum.lt( fDiff(initialSeed), fDiff( anum.plus(initialSeed,initialStep) )) )   // check that it's an increasing function

  @tailrec
  private def sweep( seed: A, step: A ): A = {
    val bDiff= fDiff(seed)

    if ( bnum.lt( bnum.abs(bDiff), bEps) ) {  // done
      seed
    } else if ( bnum.signum(bDiff) != anum.signum(step) ) {
      sweep( anum.plus(seed,step), step )   // continue, same step and direction ('bDiff' should go smaller)
    } else {
      val newStep = anum.toDouble(step) / -2.0
      sweep( anum.minus(seed,newStep), newStep )    // reverse, smaller step
    }
  }

  // Make sure we take the initial step in the right direction
  //  
  private lazy val stepSign= -bnum.signum( fDiff(initialSeed) )

  def apply: A = sweep( initialSeed, stepSign * initialStep )
}

object TestX extends App {

  val t= new Sweep( (a: Double) => (a*a)-2, 1.0, 0.5, 1e-3 )()

  println( t, math.sqrt(2.0) )
}

我也尝试过使用旧的 (implicit anum: Numeric[A]) 参数，但是无法有两个这样的参数(A 和 B).

这是编译器所说的(Scala 2.9):

fsc -deprecation -d out-make -unchecked src/xxx.scala
src/xxx.scala:25: error: type mismatch;
 found   : newStep.type (with underlying type Double)
 required: A
      sweep( anum.minus(seed,newStep), newStep )    // reverse, smaller step
                             ^
src/xxx.scala:33: error: overloaded method value * with alternatives:
  (x: Double)Double <and>
  (x: Float)Float <and>
  (x: Long)Long <and>
  (x: Int)Int <and>
  (x: Char)Int <and>
  (x: Short)Int <and>
  (x: Byte)Int
 cannot be applied to (A)
  def apply: A = sweep( initialSeed, stepSign * initialStep )
                                              ^
src/xxx.scala:38: error: not enough arguments for constructor Sweep: (implicit evidence$1: Numeric[Double], implicit evidence$2: Numeric[Double])Sweep[Double,Double].
Unspecified value parameters evidence$1, evidence$2.
  val t= new Sweep( (a: Double) => (a*a)-2, 1.0, 0.5, 1e-3 )()
         ^
three errors found

感谢您的任何想法。

Answer 1

您想使用 Fractional 而不是 Numeric。以下为我编译:

import scala.annotation.tailrec
import math.Fractional.Implicits._
import Ordering.Implicits._

class Sweep[A: Fractional, B: Fractional](fDiff: A => B, initialSeed: A, initialStep: A, bEps: B) {
  val aFractional = implicitly[Fractional[A]]

  assert(initialStep.signum > 0)
  assert(fDiff(initialSeed) < fDiff(initialSeed + initialStep))

  @tailrec
  private def sweep(seed: A, step: A): A = {
    val bDiff = fDiff(seed)
    if (bDiff.abs < bEps) {
      seed
    } else if (bDiff.signum != step.signum) {
      sweep(seed + step, step)
    } else {
      val one = aFractional.one
      val newStep = step / aFractional.fromInt(-2)
      sweep(seed - newStep, newStep)
    }
  }

  private lazy val stepSign = aFractional.fromInt(-fDiff(initialSeed).signum)
  def apply: A = sweep(initialSeed, stepSign * initialStep)
}

val sweep = new Sweep((a: Double) => (a*a)-2, 1.0, 0.5, 1e-3)
println(sweep.apply, math.sqrt(2.0))

请注意，要在 A 类型中获取类似 -2.0 的内容，您需要从 Fractional.one 手动组装它们或使用Fractional.fromInt。

另一件值得指出的事情是 math.Fractional.Implicits 和 Ordering.Implicits 的使用，这将允许您使用正常的数学语法(+、<、/等)，而不是调用 plus 和 div 等函数。