CODEWITHSUNDEEP
scala
Márcio Paiva
Márcio Paiva

Reputation: 1003

Constructor parameters with same name as atributes

In Java, I can do this:

class Point{
  int x, y;
  public Point (int x, int y){
    this.x = x;
    this.y = y;
  }
}

How can I do the same thing in Scala (use the same names in constructor arguments and in class attributes):

class Point(x: Int, y: Int){
  //Wrong code
  def x = x;
  def y = y;
}

Edit

I'm asking this because the code below doesn't work

class Point(x: Int, y: Int) {
    def +(that: Point): Point = new Point(this.x + that.x, this.y + that.y)
}

But the following one works:

class Point(px: Int, py: Int) {
  def x = px
  def y = py
  def +(that: Point): Point = new Point(this.x + that.x, this.y + that.y)
}

Upvotes: 0

Views: 2079

Answers (2)

Brian
Brian

Reputation: 20285

In Scala the parameters of the constructor become public attributes of the class if declared as a var or val.

scala> class Point(val x: Int, val y: Int){}
defined class Point

scala> val point = new Point(1,1)
point: Point = Point@1bd53074

scala> point.x
res0: Int = 1

scala> point.y
res1: Int = 1

Edit to answer the question in comments "if they were private fields, shouldn't my first code snipped after the edit have worked?"

The constructor class Point(x: Int, y: Int) generates object-private fields which only allow methods of the Point class to access the fields x and y not other objects of type Point. that in the + method is another object and is not allowed access with this definition. To see this in action define add a method def xy:Int = x + y which does not generate a compile error.

To have x and y accessible to the class use a class-private field which is as follows:

class Point(private val x: Int, private val y: Int) {
    def +(that: Point): Point = new Point(this.x + that.x, this.y + that.y)
}

Now they are not accessible outside of the class:

scala> val point = new Point(1,1)
point: Point = Point@43ba9cea

scala> point.x
<console>:10: error: value x in class Point cannot be accessed in Point
              point.x
                    ^
scala> point.y
<console>:10: error: value y in class Point cannot be accessed in Point
              point.y

You can see this in action by using scalac -Xprint:parser Point.scala.

Upvotes: 6

Ray Toal
Ray Toal

Reputation: 88428

You don't need to; the "arguments" in the class declarations are all you need in Scala.

Upvotes: 0

Related Questions