The wiki page says both classes deal with container operations, with Foldable being a class of containers that have foldr
defined over them, and for Functor it's fmap
.
However, what is the fundamental difference between types that are Foldable and types that are Functors?
The wiki hints at this distinction:
The class [Foldable] does not require Functor superclass in order to allow containers like Set or StorableVector
But I'm still not sure why a Set couldn't be mapped over to obtain another set, if I'm interpreting that correctly.
Foldable
and Functor
offer two separate abstractions for types with structures that can be folded (or reduced) and mapped over, respectively.
A Foldable holds values that can be enumerated and combined together1. One can think of a Foldable as something that can be turned into a list (toList :: Foldable f => f a -> [a]
). Alternatively, one can think of Foldables as structures whose values can be combined monoidally: (Foldable t, Monoid m) => (a -> m) -> t a -> m
(of course, this requires the ability to enumerate them).
Functors, on the other hand, are structures that allow one to "lift" a function (a -> b)
to apply to the a
s held by the structure (fmap :: (a -> b) -> (f a -> f b)
). fmap
must preserve the structure being mapped over: a tree must have the same shape before and after, a list must have the same number of elements in the same order, Nothing
can't be turned into something, and so on. On the other hand, Foldables do not need to preserve this structure; the whole point is to discard the structure and produce a new one.
The Wiki refers to the fact that there is no way to supply a typeclass constraint to fmap
. fmap :: (Ord a, Ord b) => (a -> b) -> Set a -> Set b
does not unify with the type defined by the class, fmap :: (a -> b) -> f a -> f b
, which has no constraints. This makes an instance for Set
impossible to write.
However, this is just a language implementation issue and not a deeper mathematic statement about sets. The real reason that Foldable does not have a Functor superclass is simply that there are Foldable instances which are not Functor instances.
Proxy s a
holds zero a's, Identity a
holds one a, Maybe a
holds zero or one a, b -> a
holds |b|
a's, and so on.Suppose I have xs :: Set Int
and I want to map the function putStrLn
over it. This will be a problem because IO ()
doesn't have an Ord
instance, so there's no way to figure out how to insert those actions into the result set. The type of fmap
leaves no room for constraints on the type argument to the Functor
.
IO
also provides an example of something that's a Functor
but that there isn't any meaningful way to foldMap
.
It's worth mentioning that Foldable
and Functor
come together in the Traversable
class, which gives substantially more power than either.
Set
and StorableVector
are functors, you can indeed map functions over them.
But not any kind of function. For instance, you can't map (+)
over a StorableArray
of numbers: this would give an array of functions, and these are not storable.
So, these functors are not (endo-)functors on all of Hask, but only on a subcategory including the types of a particular class. This can't be expressed in Haskell98, in fact it has only become possible quite recently with the advent of constraint kinds. See this example:
instance Functor Set Ranking Ranking where
fmap = constrainedFmap Set.map
Actually, sets also form a monad in Hask if you use some clever GADT tricks. But this isn't possible for all Foldable
containers, hence the standard library doesn't require Functor f => Foldable f
.
The Functor
typeclass doesn’t let you fmap
over things that have constraints on the types of their elements. Set
requires Ord
and StorableVector
requires Storable
.
It is possible to express a “constrained functor” using GHC’s ConstraintKinds
extension, something like:
{-# LANGUAGE ConstraintKinds,
FunctionalDependencies,
MultiParamTypeClasses #-}
import GHC.Exts (Constraint)
import Data.Set (Set)
import qualified Data.Set as Set
class ConstrainedFunctor c f | f -> c where
cfmap :: (c a, c b) => (a -> b) -> f a -> f b
instance ConstrainedFunctor Ord Set where
cfmap = Set.map
But this machinery wasn’t around when Functor
first appeared.
In addition, Functor
is supposed to be “shape-preserving”, but e.g. mapping over a Set
may change its size.
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