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larry niven Category: SF Fiction / Literature / Fantasy Category: Short story collectionsQ: scalaz broadcast - how does it work? I'm playing with the example from Scalaz docs object Broadcast { def apply[A](implicit F: Monad[F]): F[List[A]] =??? } where F is Scalaz applicative functor. So far so good. When I try to run: implicit val F = scala.math.NumericOps.scalaz.Monad[F] println(Broadcast(List(1, 2, 3, 4, 5))) I get error: Error:(19, 31) missing parameter type for expanded function ((x$1) => x$1.scalaz.Coproduct.$Monad#withUnit(Int).apply(List(1, 2, 3, 4, 5))) println(Broadcast(List(1, 2, 3, 4, 5))) If I add x: A as argument it's ok, but I do not see the point of it. Can anyone please explain how Broadcast.apply works? A: Broadcast takes a Functor. It tries to apply fmap on F, and if it succeeds, it returns F[List[A]]. The call to F[List[A]] is a Monad, so it can do the work, and all you need to do is to implement fmap. You have two options: use scalaz.Bifunctor use scalaz.functor (Bifunctor is a sub-class of Functor) The first one would look like this: def fmap[A, B](implicit b: Bifunctor[F, B]): b.A => b.B So if A and B are of the same type, you can use fmap from a Monad. The second one would look like this: def fmap[A, B](implicit e: Monad[F]): F[List[A]] => F[List[B]] Here, fmap is applied to a Monad, but the result is a Functor (because F is a Monad). You can compose F and the Monad in order to get a Functor.

 

 

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