CODEWITHSUNDEEP
scala
Irina Rapoport
Irina Rapoport

Reputation: 1692

Is there an elegant way to call a constructor with many arguments?

I am reading a CSV file with many columns and turning every row into an object. It goes like this: 

  val sighting = Sighting(
    cols(0).toInt,
    cols(1),
    cols(2),
    cols(3),
    cols(4),
    cols(5).toInt,
    cols(6),
    cols(7),
    cols(8),
    cols(9).toDouble
    cols(10),
    cols(11),
    cols(12))

This looks seriously ugly to me. Short of named parameters (which would be an improvement, granted), is there anything I can do to make it look less ugly?

Upvotes: 1

Views: 62

Answers (2)

fcat
fcat

Reputation: 1251

You can transfer the ugliness into a constructor which accepts an Array. I am aware that this doesn't reduce the level of ugliness, however still a better place to handle.

 object Sightings{

   def apply(array: Array[String]): Sightings = new Sightings(array(0), array(1), ...)

 }

Then you can use:

 val strings: Array[String] = line.split(",") //split the line and convert it into array of strings
 val sightings: Sightings = Sightings(strings) //create Sighting from the array

Upvotes: 0

蘇哲聖
蘇哲聖

Reputation: 795

How about this:

val sighting = cols match { case Seq(c0, c1, c2, ..., c12) => 
  Sighting(c0.toInt, c1, c2, ...)
}

Upvotes: 2

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