Question

我目前正在学习Scala编程语言（顺便说一句，喜欢它）并且最近发现了隐式转换。所以我有我的类Complex和它的伴随对象。在我的伴侣对象中，我希望他们隐含转换：

    implicit def convertToComplex(x: Double) = new Complex(x, 0);
    implicit def convertToComplex(x: Int) = new Complex(x, 0);
    implicit def convertToComplex(x: Float) = new Complex(x, 0);
    implicit def convertToComplex(x: Short) = new Complex(x, 0);
    implicit def convertToComplex(x: Long) = new Complex(x, 0);

为了不使用+， - ，*和/的重载来增加我的复数类，以便添加数字类型。

所以我的问题是：

是否有一种较短的方法来编写带有一些语法糖的隐式转换？（我的意思是没有所有重载...）

提前致谢。

Answer 1

你可以做这样的事情

object ConvertComplex {
  case class Complex[T: Numeric](x: T, y: Int) {

    val nX = implicitly[Numeric[T]]

    def doSomething = nX.toInt(x) * y
    def doSomethingElse = nX.plus(x, nX.fromInt(y))
  }

  object Complex {
    implicit def convertToComplect[T: Numeric](x: T) = Complex(x, 0)
  }

}

测试结果：

def doSomething[T: Numeric](x: Complex[T]) = x.doSomething
def doSomethingElse[T: Numeric](x: Complex[T]) = x.doSomething

val double = 1: Double    //> double  : Double = 1.0
val int = 1: Int          //> int  : Int = 1
val float = 1: Float      //> float  : Float = 1.0
val short = 1: Short      //> short  : Short = 1
val long = 1: Long        //> long  : Long = 1

doSomething(double)       //> res0: Int = 0
doSomething(int)          //> res1: Int = 0
doSomething(float)        //> res2: Int = 0
doSomething(short)        //> res3: Int = 0
doSomething(long)         //> res4: Int = 0

doSomethingElse(double)   //> res5: Double = 1.0
doSomethingElse(int)      //> res6: Int = 1
doSomethingElse(float)    //> res7: Float = 1.0
doSomethingElse(short)    //> res8: Short = 1
doSomethingElse(long)     //> res9: Long = 1

Answer 2

如果您希望Integral使用方法Numeric，则需要Complex而不是/。

这是一个几乎完成的解决方案：

scala> :paste
// Entering paste mode (ctrl-D to finish)

import scala.math.Integral

class Complex[T: Integral](val real: T, val imag: T) {
  import Integral.Implicits._
  def +(that: Complex[T]) = Complex(real+that.real, imag+that.imag)
  def -(that: Complex[T]) = Complex(real-that.real, imag-that.imag)
  def *(that: Complex[T]) = Complex(real*that.real - imag*that.imag, real*that.imag+imag*that.real)
  def /(that: Complex[T]) = {
    val d = that.real*that.real + that.imag*that.imag
    Complex((real*that.real + imag*that.imag)/d, (-real*that.imag+imag*that.real)/d)
  }
  override def toString: String = imag.signum match {
    case 1  => "" + real + "+" + imag +"i"
    case -1 => "" + real + "" + imag + "i"
    case _  => "" + real
  }
}

object Complex {
  import Integral.Implicits._
  def apply[T: Integral](real: T, imag: T): Complex[T] = new Complex[T](real, imag)
  object i extends Complex[Byte](0.toByte, 1.toByte)
  implicit class Real2Imag[T](val imag: T) {
    def *(v: i.type)(implicit ig: Integral[T]) = Complex(ig.zero, imag)
  }
  implicit def promote[U, V](c: Complex[U])(implicit ev: U => V, il: Integral[V]) =
    Complex(ev(c.real), ev(c.imag))
  implicit def real2complex[T: Integral](real: T) = Complex(real, implicitly[Integral[T]].zero)
  implicit val doubleIsIntegral = scala.math.Numeric.DoubleAsIfIntegral
}

// Exiting paste mode, now interpreting.

warning: there were 2 feature warnings; re-run with -feature for details
import scala.math.Integral
defined class Complex
defined object Complex

scala> import Complex._
import Complex._

scala> val c1 = 3 + 4*i
c1: Complex[Int] = 3+4i

scala> val c2 = 3.0 + 4.0*i
c2: Complex[Double] = 3.0+4.0i

scala> val c3 = c1/c2
c3: Complex[Double] = 1.0

scala> val c4 = 3 - 4*i
c4: Complex[Int] = 3-4i

scala> val c5 = c2*c4
c5: Complex[Double] = 25.0

您可以将pow方法编写为此（请注意如何生成Complex个）：

  def pow(n: Int): Complex[T] = {
    val one = Complex[T](implicitly[Integral[T]].one, implicitly[Integral[T]].zero)
    Iterator.iterate(one)(_*this).drop(n).next
  }

scala> i.pow(2)
res5: Complex[Byte] = -1

scala> i.pow(10)
res6: Complex[Byte] = -1

scala> (1+1*i).pow(4)
res7: Complex[Int] = -4

scala> (1+1*i).pow(5)
res8: Complex[Int] = -4-4i

Answer 3

是的，有。我假设Complex类有Double个参数。然后你可以一次写一个隐式转换。

implicit def doubleToComplex[T <% Double](x: T): Complex = new Complex(x, 0)

Scala隐式转换任何数字类型

3 个答案: