Why do I get a List when I create Seq using the following code?

    scala> val s = Seq[Int]()
    s: Seq[Int] = List()

In fact, Seq is a trait and trait cannot be initialized. What't the magic behind the scene

CodePudding user response：

You called apply from the Seq companion object. The implementation of apply returned a List[Int], which is an instance of Seq[Int]

CodePudding user response：

Seq is a trait, List is a (default) implementation.

object Seq extends SeqFactory.Delegate[Seq](List)
//                                          ^^^^

https://github.com/scala/scala/blob/v2.13.10/src/library/scala/collection/immutable/Seq.scala#L39

implicitly[List[Int] <:< Seq[Int]] // compiles, i.e. List[Int] is a subtype of Seq[Int]

Seq is a trait and trait cannot be initialized

Firstly, even if X is a trait you can instantiate an anonymous class extending the trait: new X {}. (By the way, List is also an abstract class.)

Secondly, Seq[Int]() is desugared into Seq.apply[Int]() and you refer to not the trait Seq but its companion object.