Scala construct objects with or without arguments

Question

In Scala, need a class that will construct a default object based on predefined computation (below the code may not be syntactically correct, but the idea is shown) if its no-params constructor is called. And I will be able to test its methods by creating an object with this(i, s) and parameters created outside. What is the best way to do that?

class myobj(i: Int, s: String) {
    def this() = {
        val j = 7 // in reality more computation with extra vals involved
        val i = j
        val str = "abcdefg"
        val s = str.get(indexOf (i % 5))
        this(i, s)
    }
}

dhg · Accepted Answer · 2015-03-11 13:21:54Z

3

It might be better with a static factory:

class MyObj(i: Int, s: String)

object MyObj {
  def apply() = {
    val j = 7 // in reality more computation with extra vals involved
    val i = j
    val str = "abcdefg"
    val s = ""
    new MyObj(i, s)
  }
}

Then you can just do:

val o = MyObj()

answered Mar 11, 2015 at 13:21

dhg

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4 Comments

Stephan Rozinsky Over a year ago

Thank you, did like you suggested. Do you know how can I then retrieve o.i and o.s. The compiler tells me value i is not a member of object

Sergii Lagutin Over a year ago

@StephanRozinsky add val to class definition: class MyObj(val i: Int, val s: String)

MeiSign Over a year ago

make MyObj a case class instead of a class and you should be able to access the parameters...

Stephan Rozinsky Over a year ago

@Sergey Lagutin дзякуй :)

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