scala 2.8 breakout

Posted by oxbow_lakes on Stack Overflow See other posts from Stack Overflow or by oxbow_lakes
Published on 2009-11-11T14:53:23Z Indexed on 2010/05/08 15:08 UTC
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In Scala 2.8, there is an object in scala.collection.package.scala:

def breakOut[From, T, To](implicit b : CanBuildFrom[Nothing, T, To]) =
  new CanBuildFrom[From, T, To] {
    def apply(from: From) = b.apply() ; def apply() = b.apply()
  }

I have been told that this results in:

> import scala.collection.breakOut
> val map : Map[Int,String] = List("London", "Paris").map(x => (x.length, x))(breakOut)

map: Map[Int,String] = Map(6 -> London, 5 -> Paris)

What is going on here? Why is breakOut being called as an argument to my List?

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