Scala weird map function behaviour
scala
Solution
By way of `collection.Seq[Char]`, which is a subtype of `PartialFunction[Int, Char]`, which is a subtype of `Int => Char`:
scala> implicitly[collection.immutable.WrappedString <:< (Int => Char)]
res0: <:<[scala.collection.immutable.WrappedString,Int => Char] = <function1>
So there's only one implicit conversion happening—the original `String => WrappedString`, which kicks in because we're treating a string like a function.
Problem
Why does the following code work? ``` scala> List(1,2,3) map "somestring" res0: List[Char] = List(o, m, e) ``` It works in both 2.9 and 2.10. Looking into the typer: ``` [master●●] % scala -Xprint:typer -e 'List(1,2,3) map "somestring"' ~/home/folone/backend [[syntax trees at end of typer]] // scalacmd2632231162205778968.scala package <empty> { object Main extends scala.AnyRef { def <init>(): Main.type = { Main.super.<init>(); () }; def main(argv: Array[String]): Unit = { val args: Array[String] = argv; { final class $anon extends scala.AnyRef { def <init>(): anonymous class $anon = { $anon.super.<init>(); () }; immutable.this.List.apply[Int](1, 2, 3).map[Char, List[Char]](scala.this.Predef.wrapString("somestring"))(immutable.this.List.canBuildFrom[Char]) }; { new $anon(); () } } } } } ``` Looks like it gets converted to the `WrappedString`, which has an apply method. This explains, how it works, but does not explain, how a `WrappedString` got accepted into a parameter of type `A => B` (as specified in the scaladoc). Can someone explain, how this happens, please?