Question

Scala’s handling of superclass constructor parameters is confusing me…

with this code:

class ArrayElement(val contents: Array[String]) {
   ...
}

class LineElement(s: String) extends ArrayElement(Array(s)) {
  ...
}

LineElement is declared to extend ArrayElement, it seems strange to me that the Array(s) parameter in ArrayElement(Array(s)) is creating an Array instance – runtime??? Is this scala’s syntatic sugar or is there something else going on here?

Answer 1

Yes, the Array(s) expression is evaluated at run-time.

class Foo (val x: Int)
class Bar (x: Int, y: Int) extends Foo(x + y)

Scala allows expressions in the calls to a superclass’ constructor (similar to what Java does with its use of super(...)). These expressions are evaluated at run-time.