如何将Scala构造函数参数公开为公共成员？

Question

看看这个例子:

class Point(x: Double, y: Double){
  override def toString = "x: " + x + ", y: " + y
  def +(sourcePoint: Point) : Point = {
    return new Point(x + sourcePoint.x, y + sourcePoint.y
  }
}

如您所见,我想+在Point类上定义一个运算符方法.但是,这不会在方法,因为工作,x而y不能在访问sourcePoint局部变量,因为它们是私有的,所以我改变的例子为这样的:

class Point(_x: Double, _y: Double){
  var x = _x
  var y = _y

  override def toString = "x: " + x + ", y: " + y
  def +(sourcePoint: Point) : Point = {
    return new Point(x + sourcePoint.x, y + sourcePoint.y)
  }
}

这显然有效,但是有一种更简单的方法来定义这些变量而不是从_x - > x和_y - > y.

感谢您的帮助和时间!:)

Answer 1

就在这里:

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

Answer 2

使用val有效,但参数变为final(常量).如果您希望能够重新分配您应该使用的值var.所以

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