Where has Seq.apply gone in Scala 2.9.2? - scala

Question

Scala 2.8.2 has a Seq.apply method, so that you can write the following in the REPL:
val l = Seq(1, 2)
l: Seq[In] = List(1, 2)
This still works in Scala 2.9.2, but what puzzles me is that according to the documentation, there is no such thing as a scala.collection.Seq.apply method anymore.
I checked with scala -Xprint:typer and this is what it prints:
[[syntax trees at end of typer]]// Scala source: <console>
package $line14 {
final object $read extends java.lang.Object with ScalaObject {
def this(): object $line14.$read = {
$read.super.this();
()
};
final object $iw extends java.lang.Object with ScalaObject {
def this(): object $line14.$read.$iw = {
$iw.super.this();
()
};
final object $iw extends java.lang.Object with ScalaObject {
def this(): object $line14.$read.$iw.$iw = {
$iw.super.this();
()
};
private[this] val l: Seq[Int] = collection.this.Seq.apply[Int](1, 2);
<stable> <accessor> def l: Seq[Int] = $iw.this.l
}
}
}
}
[[syntax trees at end of typer]]// Scala source: <console>
package $line14 {
final object $eval extends java.lang.Object with ScalaObject {
def this(): object $line14.$eval = {
$eval.super.this();
()
};
lazy private[this] var $result: Seq[Int] = {
$eval.this.$print;
$line14.$read.$iw.$iw.l
};
private[this] val $print: String = {
$read.$iw.$iw;
"l: Seq[Int] = ".+(scala.runtime.ScalaRunTime.replStringOf($line14.$read.$iw.$iw.l, 1000))
};
<stable> <accessor> def $print: String = $eval.this.$print
}
}
l: Seq[Int] = List(1, 2)
So the result is effectively:
collection.this.Seq.apply[Int](1, 2)
which indicates that it still calls Seq.apply, but where is this method?

Answer 1

It's really a bug in Scaladoc 2.9.2.
I've inspected the scala-library.jar and found that the companion object scala.collection.Seq indirectly extends scala.collection.generic.GenericCompanion[Seq], which effectively provides the method signature:
def apply[A](elems: A*): Seq[A]
This in turn calls newBuilder, which is overridden in Seq to return:
scala.collection.immutable.Seq.newBuilder[A]