9

i try to learn/understand a little bit of scalaz. For that i started with the example:

List(3, 4, 5).asMA.foldMap(x => x)
 => 12 //(3+4+5) 

def foldMap[B](f: A => B)(implicit r: Foldable[M], m: Monoid[B])

So somewhere has to be an Foldable[List[_]] and a Monoid[Int] (with append = + and zero = 0). But i wasn't able to find these two implicits. Is there an easy way to find them?

Then the next example was:

List(3, 4, 5).asMA.foldMap(multiplication)
 => 60 //(3*4*5)

Here i get even more confused. I assumed that multiplication has to be replace the Monoid[Int] with one with append = *, zero = 1. But then f: A=>B is missing. And if i follow multiplication i don't find anything connected to a Monoid or function etc.

sealed trait IntMultiplication extends NewType[Int]

trait NewType[X] {
  val value: X
  override def toString = value.toString
}
Fabian
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2 Answers2

9

There are some handy compiler flags when dealing with implicits: -Xlog-implicits, -Xprint:typer and -Ytyper-debug

In this case you can use -Xprint:typer flag to see expressions with applied implicits. Then, first snippet List(3, 4, 5).asMA.foldMap(identity) will expand to

scalaz.this.Scalaz.SeqMA[List, Int](immutable.this.List.apply[Int](3, 4, 5)).asMA.foldMap[Int]({
  ((x: Int) => scala.this.Predef.identity[Int](x))
})(scalaz.this.Foldable.ListFoldable,
   scalaz.this.Monoid.monoid[Int](scalaz.this.Semigroup.IntSemigroup, scalaz.this.Zero.IntZero));

Now it is clear that

Monoid.monoid[Int](Semigroup.IntSemigroup, Zero.IntZero)

is used to create Monoid[Int] instance (with append = + and zero = 0)

Second snippet, List(3, 4, 5).foldMap(multiplication) will expand to

scalaz.this.Scalaz.SeqMA[List, Int](immutable.this.List.apply[Int](3, 4, 5)).foldMap[scalaz.IntMultiplication]({
  ((n: Int) => scalaz.Scalaz.multiplication(n))
})(scalaz.this.Foldable.ListFoldable,
   scalaz.this.Monoid.monoid[scalaz.IntMultiplication](scalaz.this.Semigroup.IntMultiplicationSemigroup, scalaz.this.Zero.IntMultiplicationZero));

In this case Monoid[IntMultiplication] (with append = * and zero = 1) is used as implicit parameter.

Update

To create Monoid for your type, you need to have implicit Semigroup and Zero in scope

case class Foo(x: Int)

implicit def FooSemigroup: Semigroup[Foo] = semigroup((f1, f2) => Foo(f1.x + f2.x))
implicit def FooZero: Zero[Foo] = zero(Foo(0))

scala> (1 to 10) map Foo foldMap identity
res5: Foo = Foo(55)
4e6
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  • When i would want to add a Monoid for a type i have. I would only need an implicit def for Zero and for Semigroup? – Fabian May 01 '12 at 11:35
  • Yes, you need implicit `Semigroup[YourType]` and `Zero[YourType]` in scope – 4e6 May 01 '12 at 11:59
7

Scalaz 6 stores the typeclass instances in the companion objects of the typeclasses themselves, so you need to be looking there.

In the case of Monoid[T], the actual typeclass is splittable into two sections: Semigroup[T] provides us with the append operation (T, T) => T, Zero[T] provides us with a zero function: () => T. The actual Monoid instance can then be generated via the implicit def monoid living inside MonoidLow.

For foldable, the typeclass for Foldable[List] lives inside the Foldable singleton and is called ListFoldable.

As I'm sure you've worked out, the default monoid in scope for integers is (+, 0), so the multiplication function you've used in the second example is just a converter to the wrapper type IntMultiplication, which has semigroup and zero instances defined inside the relevant companion objects.

If you're trying to get your head around Scalaz, there's a couple of good introduction videos. The ones that I found most useful were these two:

[1] http://skillsmatter.com/podcast/scala/practical-scalaz-2518/js-1603

[2] http://www.infoq.com/presentations/Scalaz-Functional-Programming-in-Scala#.T0q_hgmiJbw.wordpress

Both of which touch on monoids.

AlecZorab
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