CODEWITHSUNDEEP
scalascala-collections
Beryllium
Beryllium

Reputation: 12988

Why is Seq.newBuilder returning a ListBuffer?

Looking at

val sb = Seq.newBuilder[Int]
println(sb.getClass.getName)
sb += 1
sb += 2
val s = sb.result()
println(s.getClass.getName)

the output is

scala.collection.mutable.ListBuffer
scala.collection.immutable.$colon$colon

using Scala 2.10.1.

I would expect Seq.newBuilder to return a VectorBuilder for example. This is returned by CanBuildFrom, if the result is explicitly typed to a Seq:

def build[T, C <: Iterable[T]](x: T, y: T)
                              (implicit cbf: CanBuildFrom[Nothing, T, C]): C = {
  val b = cbf()
  println(b.getClass.getName)
  b += x
  b += y
  b.result()
}

val s: Seq[Int] = build(1, 2)
println(s.getClass.getName) // scala.collection.immutable.Vector

in this case the builder is a VectorBuilder, and the result's class is a Vector.

So I explicitly wanted to build a Seq, but the result is a List which needs more RAM, according to Scala collection memory footprint characteristics.

So why does Seq.newBuilder return a ListBuffer which gives a List in the end?

Upvotes: 4

Views: 5220

Answers (4)

som-snytt
som-snytt

Reputation: 39577

OK, but you're not going to believe it. Turning on -Yinfer-debug for your CanBuildFrom counter-example,

[search] $line14.$read.$iw.$iw.build[scala.this.Int, Seq[scala.this.Int]](1, 2) with pt=generic.this.CanBuildFrom[scala.this.Nothing,scala.this.Int,Seq[scala.this.Int]] in module class $iw, eligible:
  fallbackStringCanBuildFrom: [T]=> generic.this.CanBuildFrom[String,T,immutable.this.IndexedSeq[T]]
[solve types] solving for T in ?T
inferExprInstance {
  tree      scala.this.Predef.fallbackStringCanBuildFrom[T]
  tree.tpe  generic.this.CanBuildFrom[String,T,immutable.this.IndexedSeq[T]]
  tparams   type T
  pt        generic.this.CanBuildFrom[scala.this.Nothing,scala.this.Int,Seq[scala.this.Int]]
  targs     scala.this.Int
  tvars     =?scala.this.Int
}
[search] considering no tparams (pt contains no tvars) trying generic.this.CanBuildFrom[String,scala.this.Int,immutable.this.IndexedSeq[scala.this.Int]] against pt=generic.this.CanBuildFrom[scala.this.Nothing,scala.this.Int,Seq[scala.this.Int]]
[success] found SearchResult(scala.this.Predef.fallbackStringCanBuildFrom[scala.this.Int], ) for pt generic.this.CanBuildFrom[scala.this.Nothing,scala.this.Int,Seq[scala.this.Int]]
[infer implicit] inferred SearchResult(scala.this.Predef.fallbackStringCanBuildFrom[scala.this.Int], )

and indeed,

  implicit def fallbackStringCanBuildFrom[T]: CanBuildFrom[String, T, immutable.IndexedSeq[T]] =
    new CanBuildFrom[String, T, immutable.IndexedSeq[T]] {
      def apply(from: String) = immutable.IndexedSeq.newBuilder[T]
      def apply() = immutable.IndexedSeq.newBuilder[T]
    }

What do you mean, your Iterable is not a String?

trait CanBuildFrom[-From, -Elem, +To]

Such is the evil of inferring either Nothing or Any.

Edit: Sorry, I misspoke, I see that you told it Nothing explicitly.

Update:

Since CBF is contravariant in From, a CBF from String serves as a CBF from Nothing.

scala> typeOf[CanBuildFrom[Nothing,Int,Seq[Int]]] <:< typeOf[CanBuildFrom[String,Int,Seq[Int]]]
res0: Boolean = false

scala> typeOf[CanBuildFrom[String,Int,Seq[Int]]] <:< typeOf[CanBuildFrom[Nothing,Int,Seq[Int]]]
res1: Boolean = true

For instance, if you need to build from an immutable.Map, you'd want a CBF from collection.Map to work.

As someone else commented, it's just weird for Nothing. But you get what you asked for. That is, you underspecified, which means you don't mind much what you get back, Vector or whatever.

Upvotes: 1

0__
0__

Reputation: 67280

The default Seq implementation is List:

Seq(1, 2, 3)  // -> List(1, 2, 3)

...thus ListBuffer is the correct builder. If you want Vector, use Vector.newBuilder or IndexedSeq.newBuilder.

Upvotes: 4

Fynn
Fynn

Reputation: 4873

The Scala Collections API is very complex and its hierarchy is rich in depth. Each level represents some sort of new abstraction. The Seq trait split up into two different subtraits, which give different guarantees for performance (ref.):

  1. An IndexedSeq provides fast random-access of elements and a fast length operation. One representative of this IndexedSeq is the Vector.

  2. A LinearSeq provides fast access only to the first element via head, but also has a fast tail operation. One representative of this LinearSeq is the List.

As the current default implementation of a Seq is a List, Seq.newBuilder will return a ListBuffer. However, if you want to use a Vector you can either use Vector.newBuilder[T] or IndexedSeq.newBuilder[T]:

scala> scala.collection.immutable.IndexedSeq.newBuilder[Int]
res0: scala.collection.mutable.Builder[Int,scala.collection.immutable.IndexedSeq[Int]] = scala.collection.immutable.VectorBuilder@1fb10a9f

scala> scala.collection.immutable.Vector.newBuilder[Int]
res1: scala.collection.mutable.Builder[Int,scala.collection.immutable.Vector[Int]] = scala.collection.immutable.VectorBuilder@3efe9969

Upvotes: 8

Eve Freeman
Eve Freeman

Reputation: 33145

I agree that this is weird. Why don't you just use Vector.newBuilder, if that's what you're looking for?

scala> val sb = Vector.newBuilder[Int]
sb: scala.collection.mutable.Builder[Int,scala.collection.immutable.Vector[Int]] = scala.collection.immutable.VectorBuilder@1fb7482a

scala> println(sb.getClass.getName)
scala.collection.immutable.VectorBuilder

scala> sb += 1
res1: sb.type = scala.collection.immutable.VectorBuilder@1fb7482a

scala> sb += 2
res2: sb.type = scala.collection.immutable.VectorBuilder@1fb7482a

scala> val s = sb.result()
s: scala.collection.immutable.Vector[Int] = Vector(1, 2)

scala> println(s.getClass.getName)
scala.collection.immutable.Vector

Upvotes: 0

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