scala - 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 expect seq.newbuilder return vectorbuilder example. returned canbuildfrom, if result explicitly typed 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 case builder vectorbuilder, , result's class vector.

so explicitly wanted build seq, result list needs more ram, according scala collection memory footprint characteristics.

so why seq.newbuilder return listbuffer gives list in end?

the scala collections api complex , hierarchy rich in depth. each level represents sort of new abstraction. seq trait split 2 different subtraits, give different guarantees performance (ref.):

1. an indexedseq provides fast random-access of elements , fast length operation. 1 representative of indexedseq vector.

2. a linearseq provides fast access first element via head, has fast tail operation. 1 representative of linearseq list.

as current default implementation of seq list, seq.newbuilder return listbuffer. however, if want use vector 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