Compile error when adding List as parameter

Question

This code :

package neuralnetwork

object hopfield {
  println("Welcome to the Scala worksheet")

  object Neuron {
    def apply() = new Neuron(0, 0, false, Nil, "")
    def apply(l : List[Neuron]) = new Neuron(0, 0, false, l, "")
  }

  case class Neuron(w: Double, tH: Double, var fired: Boolean, in: List[Neuron], id: String)

  val n2 = Neuron
  val n3 = Neuron
  val n4 = Neuron
  val l = List(n2,n3,n4)
  val n1 = Neuron(l)



}

causes compile error :

type mismatch; found : List[neuralnetwork.hopfield.Neuron.type] required: List[neuralnetwork.hopfield.Neuron]

at line : val n1 = Neuron(l)

Why should this occur ? What is incorrect about implementation that is preventing the List being added ?

Ende Neu · Accepted Answer · 2014-08-24 12:03:07Z

4

You are passing in the type, n2, n3 and n4 have type Neuron.type, try adding the parenthesis:

val n2 = Neuron()
val n3 = Neuron()
val n4 = Neuron()
val l = List(n2,n3,n4)
val n1 = Neuron(l)

The difference is that with the parenthesis you actually get a Neuron class (you call the apply method) and the type will be Neuron instead of Neuron.type.

Edit:

The .type notation is called singleton type, it denotes just the object represented by the class, in this case Neuron.type returns just that singleton object, more infos are in this paper by Odersky about the scala overview at page 9.

edited Aug 24, 2014 at 12:03

answered Aug 20, 2014 at 8:58

Ende Neu

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5 Comments

Ende Neu Over a year ago

Yes, otherwise call the apply manually as Gabriele said (I like parenthesis better).

Gabriele Petronella Over a year ago

I like the parenthesis better too, actually. It's worth saying that since apply takes no parameter, the compiler needs to choose between Neuron.apply and Neuron as a type when it sees Neuron, and it will always choose the latter, since otherwise you would have no way of referring to the type of Neuron.

blue-sky Over a year ago

@Ende Neu is .type a method ? I cannot seem to find its use in Scala API?

Ende Neu Over a year ago

@blue-sky actually is not, but there's not much around, the best I could find is this post.

Ende Neu Over a year ago

@blue-sky I found out something about the .type notation and made an edit to my answer, something more detailed can be found in that paper (you don't have to read through it if you're only interested in the singleton type notation).

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