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如何在Scala中定义赋值运算符?

operators operator-overloading scala

3

CyberPlayerOne · 技术社区 · 7 年前

例如,我定义了 +* 以下包装器类中的加权和运算符 Double :

  class DoubleOps(val double: Double) {
    def +*(other: DoubleOps, weight: Double): DoubleOps =
      new DoubleOps(this.double + other.double * weight)
  }

  object DoubleOps {
    implicit def DoubleToDoubleOps(double: Double) = new DoubleOps(double)
  }

根据此定义,我可以进行以下计算:

var db: DoubleOps = 1.5
import DoubleOps._
db = db +* (2.0, 0.5)
println(db)

有什么方法可以计算 db 使用赋值运算符获得结果,如定义 +*= ,以便我可以使用:

db +*= (2.0, 0.5)

这可能吗?

1 回复 | 直到 7 年前

1

4

sarveshseri 7 年前

import scala.languageFeature.implicitConversions

class DoubleOps(var double: Double) {
  def +*(other: DoubleOps, weight: Double): DoubleOps =
    new DoubleOps(this.double + other.double * weight)

  def +*=(other: DoubleOps, weight: Double): Unit = {
    this.double = this.double + other.double * weight
  }
}

object DoubleOps {
  implicit def DoubleToDoubleOps(double: Double) = new DoubleOps(double)
}

val d: DoubleOps = 1.5

d +*= (2.0, 0.5)