有斯卡拉

Question

參數語法糖我有操作有點語法糖，一個物體上取決於：

case class EllipticOperand (p : Point) 
{ 
    def + (q : => Point) = curve.sum(p,q) 
    def * (n : => BigInt) = curve.times(p,n) 
} 
implicit def PointToOperand(p : Point) = EllipticOperand(p) 

case class EllipticMultiplier (n : BigInt) 
{ 
def * (p : => Point) = curve.times(p,n) 
} 
implicit def BigIntToOperand (n : BigInt) = EllipticMultiplier(n) 

我想在一些class SyntacticSugar[Point](curve : main.Curve[Point])封裝在其他類定義來使用它，而無需拷貝/粘貼它。

我試圖用這種方式：

val sugar = new util.SyntacticSugar(curve) 
import sugar._ 

然而，這是不行的，以後我不能用+和*。

Answer 1

如果我實現它，你認爲它只是工作，你可以在Point使用+和*作爲運營商的方式......

case class Point(x: Int, y: Int) 

trait Curve[T] { 
    def sum(p: T, q: T): T 
    def times(p: T, n: Int): T 
} 

// dummy implementation 
class PointCurve extends Curve[Point] { 
    override def sum(p: Point, q: Point) = Point(p.x+q.x, p.y+q.y) 
    override def times(p: Point, n: Int) = Point(p.x*n, p.y*n) 
} 


object util { 
    class SyntacticSugar[T](curve: Curve[T]){ 
    case class EllipticOperand(p: T){ 
     def +(q: =>T) = curve.sum(p, q) 
     def *(n: =>Int) = curve.times(p,n) 
    } 
    implicit def point2Operand(p: T) = EllipticOperand(p) 
    } 
} 

現在：

scala> val sugar = new util.SyntacticSugar(new PointCurve) 
sugar: util.SyntacticSugar[Point] = util$SyntacticSugar@4ed4b486 

scala> import sugar._ 
import sugar._ 

scala> Point(1,2) + Point(2,3) 
res0: Point = Point(3,5) 

scala> Point(1,2) * 3 
res1: Point = Point(3,6)