Scala 2.8 breakOut

Question

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?

Answer 1

I'd like to build upon Daniel's answer. It was very thorough, but as noted in the comments, it doesn't explain what breakout does.

Taken from Re: Support for explicit Builders (2009-10-23), here is what I believe breakout does:

It gives the compiler a suggestion as to which Builder to choose implicitly (essentially it allows the compiler to choose which factory it thinks fits the situation best.)

For example, see the following:

scala> import scala.collection.generic._
import scala.collection.generic._

scala> import scala.collection._
import scala.collection._

scala> import scala.collection.mutable._
import scala.collection.mutable._

scala>

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()
     |    }
breakOut: [From, T, To]
     |    (implicit b: scala.collection.generic.CanBuildFrom[Nothing,T,To])
     |    java.lang.Object with
     |    scala.collection.generic.CanBuildFrom[From,T,To]

scala> val l = List(1, 2, 3)
l: List[Int] = List(1, 2, 3)

scala> val imp = l.map(_ + 1)(breakOut)
imp: scala.collection.immutable.IndexedSeq[Int] = Vector(2, 3, 4)

scala> val arr: Array[Int] = l.map(_ + 1)(breakOut)
imp: Array[Int] = Array(2, 3, 4)

scala> val stream: Stream[Int] = l.map(_ + 1)(breakOut)
stream: Stream[Int] = Stream(2, ?)

scala> val seq: Seq[Int] = l.map(_ + 1)(breakOut)
seq: scala.collection.mutable.Seq[Int] = ArrayBuffer(2, 3, 4)

scala> val set: Set[Int] = l.map(_ + 1)(breakOut)
seq: scala.collection.mutable.Set[Int] = Set(2, 4, 3)

scala> val hashSet: HashSet[Int] = l.map(_ + 1)(breakOut)
seq: scala.collection.mutable.HashSet[Int] = Set(2, 4, 3)

You can see the return type is implicitly chosen by the compiler to best match the expected type. Depending on how you declare the receiving variable, you get different results.

The following would be an equivalent way to specify a builder. Note in this case, the compiler will infer the expected type based on the builder's type:

scala> def buildWith[From, T, To](b : Builder[T, To]) =
     |    new CanBuildFrom[From, T, To] {
     |      def apply(from: From) = b ; def apply() = b
     |    }
buildWith: [From, T, To]
     |    (b: scala.collection.mutable.Builder[T,To])
     |    java.lang.Object with
     |    scala.collection.generic.CanBuildFrom[From,T,To]

scala> val a = l.map(_ + 1)(buildWith(Array.newBuilder[Int]))
a: Array[Int] = Array(2, 3, 4)

Answer 2

A simple example to understand what breakOut does:

scala> import collection.breakOut
import collection.breakOut

scala> val set = Set(1, 2, 3, 4)
set: scala.collection.immutable.Set[Int] = Set(1, 2, 3, 4)

scala> set.map(_ % 2)
res0: scala.collection.immutable.Set[Int] = Set(1, 0)

scala> val seq:Seq[Int] = set.map(_ % 2)(breakOut)
seq: Seq[Int] = Vector(1, 0, 1, 0) // map created a Seq[Int] instead of the default Set[Int]

Answer 3

Daniel Sobral's answer is great, and should be read together with Architecture of Scala Collections (Chapter 25 of Programming in Scala).

I just wanted to elaborate on why it is called breakOut:

Why is it called breakOut?

Because we want to break out of one type and into another:

Break out of what type into what type? Lets look at the map function on Seq as an example:

Seq.map[B, That](f: (A) -> B)(implicit bf: CanBuildFrom[Seq[A], B, That]): That

If we wanted to build a Map directly from mapping over the elements of a sequence such as:

val x: Map[String, Int] = Seq("A", "BB", "CCC").map(s => (s, s.length))

The compiler would complain:

error: type mismatch;
found   : Seq[(String, Int)]
required: Map[String,Int]

The reason being that Seq only knows how to build another Seq (i.e. there is an implicit CanBuildFrom[Seq[_], B, Seq[B]] builder factory available, but there is NO builder factory from Seq to Map).

In order to compile, we need to somehow breakOut of the type requirement, and be able to construct a builder that produces a Map for the map function to use.

As Daniel has explained, breakOut has the following signature:

def breakOut[From, T, To](implicit b: CanBuildFrom[Nothing, T, To]): CanBuildFrom[From, T, To] =
    // can't just return b because the argument to apply could be cast to From in b
    new CanBuildFrom[From, T, To] {
      def apply(from: From) = b.apply()
      def apply()           = b.apply()
    }

Nothing is a subclass of all classes, so any builder factory can be substituted in place of implicit b: CanBuildFrom[Nothing, T, To]. If we used the breakOut function to provide the implicit parameter:

val x: Map[String, Int] = Seq("A", "BB", "CCC").map(s => (s, s.length))(collection.breakOut)

It would compile, because breakOut is able to provide the required type of CanBuildFrom[Seq[(String, Int)], (String, Int), Map[String, Int]], while the compiler is able to find an implicit builder factory of type CanBuildFrom[Map[_, _], (A, B), Map[A, B]], in place of CanBuildFrom[Nothing, T, To], for breakOut to use to create the actual builder.

Note that CanBuildFrom[Map[_, _], (A, B), Map[A, B]] is defined in Map, and simply initiates a MapBuilder which uses an underlying Map.

Hope this clears things up.

Answer 4

The answer is found on the definition of map:

def map[B, That](f : (A) => B)(implicit bf : CanBuildFrom[Repr, B, That]) : That 

Note that it has two parameters. The first is your function and the second is an implicit. If you do not provide that implicit, Scala will choose the most specific one available.

About breakOut

So, what's the purpose of breakOut? Consider the example given for the question, You take a list of strings, transform each string into a tuple (Int, String), and then produce a Map out of it. The most obvious way to do that would produce an intermediary List[(Int, String)] collection, and then convert it.

Given that map uses a Builder to produce the resulting collection, wouldn't it be possible to skip the intermediary List and collect the results directly into a Map? Evidently, yes, it is. To do so, however, we need to pass a proper CanBuildFrom to map, and that is exactly what breakOut does.

Let's look, then, at the definition of breakOut:

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()
  }

Note that breakOut is parameterized, and that it returns an instance of CanBuildFrom. As it happens, the types From, T and To have already been inferred, because we know that map is expecting CanBuildFrom[List[String], (Int, String), Map[Int, String]]. Therefore:

From = List[String]
T = (Int, String)
To = Map[Int, String]

To conclude let's examine the implicit received by breakOut itself. It is of type CanBuildFrom[Nothing,T,To]. We already know all these types, so we can determine that we need an implicit of type CanBuildFrom[Nothing,(Int,String),Map[Int,String]]. But is there such a definition?

Let's look at CanBuildFrom's definition:

trait CanBuildFrom[-From, -Elem, +To] 
extends AnyRef

So CanBuildFrom is contra-variant on its first type parameter. Because Nothing is a bottom class (ie, it is a subclass of everything), that means any class can be used in place of Nothing.

Since such a builder exists, Scala can use it to produce the desired output.

About Builders

A lot of methods from Scala's collections library consists of taking the original collection, processing it somehow (in the case of map, transforming each element), and storing the results in a new collection.

To maximize code reuse, this storing of results is done through a builder (scala.collection.mutable.Builder), which basically supports two operations: appending elements, and returning the resulting collection. The type of this resulting collection will depend on the type of the builder. Thus, a List builder will return a List, a Map builder will return a Map, and so on. The implementation of the map method need not concern itself with the type of the result: the builder takes care of it.

On the other hand, that means that map needs to receive this builder somehow. The problem faced when designing Scala 2.8 Collections was how to choose the best builder possible. For example, if I were to write Map('a' -> 1).map(_.swap), I'd like to get a Map(1 -> 'a') back. On the other hand, a Map('a' -> 1).map(_._1) can't return a Map (it returns an Iterable).

The magic of producing the best possible Builder from the known types of the expression is performed through this CanBuildFrom implicit.

About CanBuildFrom

To better explain what's going on, I'll give an example where the collection being mapped is a Map instead of a List. I'll go back to List later. For now, consider these two expressions:

Map(1 -> "one", 2 -> "two") map Function.tupled(_ -> _.length)
Map(1 -> "one", 2 -> "two") map (_._2)

The first returns a Map and the second returns an Iterable. The magic of returning a fitting collection is the work of CanBuildFrom. Let's consider the definition of map again to understand it.

The method map is inherited from TraversableLike. It is parameterized on B and That, and makes use of the type parameters A and Repr, which parameterize the class. Let's see both definitions together:

The class TraversableLike is defined as:

trait TraversableLike[+A, +Repr] 
extends HasNewBuilder[A, Repr] with AnyRef

def map[B, That](f : (A) => B)(implicit bf : CanBuildFrom[Repr, B, That]) : That 

To understand where A and Repr come from, let's consider the definition of Map itself:

trait Map[A, +B] 
extends Iterable[(A, B)] with Map[A, B] with MapLike[A, B, Map[A, B]]

Because TraversableLike is inherited by all traits which extend Map, A and Repr could be inherited from any of them. The last one gets the preference, though. So, following the definition of the immutable Map and all the traits that connect it to TraversableLike, we have:

trait Map[A, +B] 
extends Iterable[(A, B)] with Map[A, B] with MapLike[A, B, Map[A, B]]

trait MapLike[A, +B, +This <: MapLike[A, B, This] with Map[A, B]] 
extends MapLike[A, B, This]

trait MapLike[A, +B, +This <: MapLike[A, B, This] with Map[A, B]] 
extends PartialFunction[A, B] with IterableLike[(A, B), This] with Subtractable[A, This]

trait IterableLike[+A, +Repr] 
extends Equals with TraversableLike[A, Repr]

trait TraversableLike[+A, +Repr] 
extends HasNewBuilder[A, Repr] with AnyRef

If you pass the type parameters of Map[Int, String] all the way down the chain, we find that the types passed to TraversableLike, and, thus, used by map, are:

A = (Int,String)
Repr = Map[Int, String]

Going back to the example, the first map is receiving a function of type ((Int, String)) => (Int, Int) and the second map is receiving a function of type ((Int, String)) => String. I use the double parenthesis to emphasize it is a tuple being received, as that's the type of A as we saw.

With that information, let's consider the other types.

map Function.tupled(_ -> _.length):
B = (Int, Int)

map (_._2):
B = String

We can see that the type returned by the first map is Map[Int,Int], and the second is Iterable[String]. Looking at map's definition, it is easy to see that these are the values of That. But where do they come from?

If we look inside the companion objects of the classes involved, we see some implicit declarations providing them. On object Map:

implicit def  canBuildFrom [A, B] : CanBuildFrom[Map, (A, B), Map[A, B]]  

And on object Iterable, whose class is extended by Map:

implicit def  canBuildFrom [A] : CanBuildFrom[Iterable, A, Iterable[A]]  

These definitions provide factories for parameterized CanBuildFrom.

Scala will choose the most specific implicit available. In the first case, it was the first CanBuildFrom. In the second case, as the first did not match, it chose the second CanBuildFrom.

Back to the Question

Let's see the code for the question, List's and map's definition (again) to see how the types are inferred:

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

sealed abstract class List[+A] 
extends LinearSeq[A] with Product with GenericTraversableTemplate[A, List] with LinearSeqLike[A, List[A]]

trait LinearSeqLike[+A, +Repr <: LinearSeqLike[A, Repr]] 
extends SeqLike[A, Repr]

trait SeqLike[+A, +Repr] 
extends IterableLike[A, Repr]

trait IterableLike[+A, +Repr] 
extends Equals with TraversableLike[A, Repr]

trait TraversableLike[+A, +Repr] 
extends HasNewBuilder[A, Repr] with AnyRef

def map[B, That](f : (A) => B)(implicit bf : CanBuildFrom[Repr, B, That]) : That 

The type of List("London", "Paris") is List[String], so the types A and Repr defined on TraversableLike are:

A = String
Repr = List[String]

The type for (x => (x.length, x)) is (String) => (Int, String), so the type of B is:

B = (Int, String)

The last unknown type, That is the type of the result of map, and we already have that as well:

val map : Map[Int,String] =

So,

That = Map[Int, String]

That means breakOut must, necessarily, return a type or subtype of CanBuildFrom[List[String], (Int, String), Map[Int, String]].