Applicative transformer classes - haskell

Where are the Applicative transformer classes? I wanted to use transformer classes for the applicative transformer stack in a previous answer, but they don't seem to exist.
The transformers package and many others are full of transformers that preserver Applicative structure, even when the underlying structure isn't a Monad.
A quick glance at transformers has Applicative instances for most of the transformers.
Applicative f => Applicative (Backwards f)
Applicative f => Applicative (Lift f)
Applicative (ContT r m)
Applicative m => Applicative (IdentityT m)
Applicative m => Applicative (ReaderT r m)
(Monoid w, Applicative m) => Applicative (WriterT w m)
(Applicative f, Applicative g) => Applicative (Compose f g)
(Applicative f, Applicative g) => Applicative (Product f g)
Only transformers for state and alternation (ExceptT and MaybeT) require an underlying monad for the Applicative instance.
(Functor m, Monad m) => Applicative (ExceptT e m)
(Functor m, Monad m) => Applicative (MaybeT m)
(Monoid w, Functor m, Monad m) => Applicative (RWST r w s m)
(Functor m, Monad m) => Applicative (StateT s m)
There's a class for Monad transformers. I can see how something could require this Monad constraint, since it can't be introduced elsewhere.
class MonadTrans t where
lift :: (Monad m) => m a -> t m a
Where's the class for Applicative transformers?
class ApTrans t where
liftAp :: (Applicative f) => f a -> t f a
Or just plain old transformers (though I can't imagine any laws for this)?
class Trans t where
liftAny :: f a -> t f a
Due to the difference only in polymorphic constraints, these typeclasses have a strange variance pattern. Except for their laws, which have to consider unexpressible constraints, anything that is an instance of Trans should automatically be an instance of ApTrans and MonadTrans, and anything that's an instance of ApTrans should automatically be an instance of MonadTrans.
If we move on to the mtl library, the classes there are also incompatible with an Applicative transformer stack. All of the mtl classes I'm familiar with have a Monad constraint. For example, here's MonadReader
class Monad m => MonadReader r m | m -> r where
-- | Retrieves the monad environment.
ask :: m r
ask = reader id
-- | Executes a computation in a modified environment.
local :: (r -> r) -- ^ The function to modify the environment.
-> m a -- ^ #Reader# to run in the modified environment.
-> m a
-- | Retrieves a function of the current environment.
reader :: (r -> a) -- ^ The selector function to apply to the environment.
-> m a
reader f = do
r <- ask
return (f r)
What is the purpose of the Monad constraint? It makes MonadReader and the MonadReader instances for many of the above transformers incompatible with Applicative transformer stacks.
I would naively write something like
class Reader r m | m -> r where
ask :: m r
local :: (r -> r) -> m a -> m a
or even split local into a separate class.
class Reader r m | m -> r where
ask :: m r
class (Reader r m) => Local r m | m -> r where
local :: (r -> r) -> m a -> m a
local might be quite hard to use without a Monad instance. A more useful interface without the Monad constraint would be something like
class (Reader r m) => Local r m | m -> r where
local :: m (r -> r) -> m a -> m a
Are there existing transformer classes somewhere that don't have the Monad constraint, or is there an actual need for yet another transformer class library?

Applicatives, unlike Monads, are closed under products and composition and thus don't need a special class of things like "transformers". Here's a small library:
data (*) f g x = P (f x) (g x) deriving Functor
data C f g x = C (f (g x)) deriving Functor
instance (Applicative f, Applicative g) => Applicative (f * g) where
pure a = P (pure a) (pure a)
P ff gf <*> P fx gx = P (ff <*> fx) (gf <*> gx)
instance (Applicative f, Applicative g) => Applicative (C f g) where
pure = C . pure . pure
C fgf <*> C fgx = C (liftA2 (<*>) fgf fgx)
Moreover, all monads are Applicatives so we ought to be able to reuse that code. Sadly, the lack of Applicative-Monad subtyping forces monadic code to be more exclusionary than needed and thus outlaws such code. It could have been rectified if all of these libraries asked for an (Applicative m, Monad m) constraint, but they do not. Furthermore, given how often you might otherwise have to write
(MonadReader m, Monad m) => ...
the Monad superclass constraint is convenient. I'm not sure it's completely necessary however.

As J. Abrahamson said, Applicatives are closed under products and composition, so there's no need for dedicated transformer versions. However, there's also no need to roll your own Applicative product/composition types, because the Platform already has these:
Data.Functor.Compose
Data.Functor.Product
Data.Functor.Constant
Data.Functor.Identity
Control.Applicative.Lift
I've found that the easier way to use these is with the GeneralizedNewtypeDeriving extension, because then you can just define types like these:
newtype MyType m a = MyType (Compose (Const m) (Reader m) a)
deriving (Functor, Applicative)
-- Plus a bunch of utility definitions to hide the use of Compose and generally
-- keep you sane...
Another other useful tool in the Applicative toolset is the free applicative functor. I normally use Edward Kmett's free library's version, but it's easy to roll your own if you want fewer dependencies.
These definitions can also be useful (though I'd welcome suggestions on the naming scheme, particularly the "I/O" bit):
{-# LANGUAGE Rank2Types, TypeOperators #-}
import Control.Applicative
import Data.Functor.Compose
-- | A handy infix type synonym for 'Compose', which allows us to
-- stack 'Applicative's with less syntactic noise:
--
-- > type CalculationT s p f = Reader (Frame s p) :. Reader (Cell s p) :. f
-- > type Calculation s p = Calculation s p Identity
--
-- Note that 'Identity' and ':.' form something a type-level monoid
-- modulo #newtype# equivalence. The following isomorphisms hold:
--
-- > f :. Identity ~= Identity :. f ~= f
-- > f :. g :. h ~= (f :. g) :. h
--
type f :. g = Compose f g
infixr :.
-- | Lift an action from the outer functor into the composite.
-- Alternative reading: append an 'Applicative' to the right of #f#.
liftO :: (Functor f, Applicative g) => f a -> (f :. g) a
liftO = Compose . fmap pure
-- | Lift an action from the inner functor into the composite.
-- Alternative reading: prepend an 'Applicative' to the left of #g#.
liftI :: Applicative f => g a -> (f :. g) a
liftI = Compose . pure
-- | Lift a natural transformation from #g# to #h# into a morphism
-- from #f :. g# to #h :. g#.
hoistO :: (forall x. f x -> h x) -> (f :. g) a -> (h :. g) a
hoistO eta = Compose . eta . getCompose
-- | Lift a natural transformation from #g# to #h# into a morphism
-- from #f :. g# to #f :. h#.
hoistI :: Functor f => (forall x. g x -> h x) -> (f :. g) a -> (f :. h) a
hoistI eta = Compose . fmap eta . getCompose

Related

Are applicative transformers really superfluous?

There is a lot of talk about Applicative not needing its own transformer class, like this:
class AppTrans t where
liftA :: Applicative f => f a -> t f a
But I can define applicative transformers that don't seem to be compositions of applicatives! For example sideeffectful streams:
data MStream f a = MStream (f (a, MStream f a))
Lifting just performs the side effect at every step:
instance AppTrans MStream where
liftA action = MStream $ (,) <$> action <*> pure (liftA action)
And if f is an applicative, then MStream f is as well:
instance Functor f => Functor (MStream f) where
fmap fun (MStream stream) = MStream $ (\(a, as) -> (fun a, fmap fun as)) <$> stream
instance Applicative f => Applicative (MStream f) where
pure = liftA . pure
MStream fstream <*> MStream astream = MStream
$ (\(f, fs) (a, as) -> (f a, fs <*> as)) <$> fstream <*> astream
I know that for any practical purposes, f should be a monad:
joinS :: Monad m => MStream m a -> m [a]
joinS (MStream stream) = do
(a, as) <- stream
aslist <- joinS as
return $ a : aslist
But while there is a Monad instance for MStream m, it's inefficient. (Or even incorrect?) The Applicative instance is actually useful!
Now note that usual streams arise as special cases for the identity functor:
import Data.Functor.Identity
type Stream a = MStream Identity a
But the composition of Stream and f is not MStream f! Rather, Compose Stream f a is isomorphic to Stream (f a).
I'd like to know whether MStream is a composition of any two applicatives.
Edit:
I'd like to offer a category theoretic viewpoint. A transformer is a "nice" endofunctor t on the category C of applicative functors (i.e. lax monoidal functors with strength), together with a natural transformation liftA from the identity on C to t. The more general question is now what useful transformers exist that are not of the form "compose with g" (where g is an applicative). My claim is that MStream is one of them.
Great question! I believe there are two different parts of this question:
Composing existing applicatives or monads into more complex ones.
Constructing all applicatives/monads from some given starting set.
Ad 1.: Monad transformers are essential for combining monads. Monads don't compose directly. It seems that there needs to be an extra bit of information provided by monad transformers that tells how each monad can be composed with other monads (but it could be this information is already somehow present, see Is there a monad that doesn't have a corresponding monad transformer?).
On the other hand, applicatives compose directly, see Data.Functor.Compose. This is why don't need applicative transformers for composition. They're also closed under product (but not coproduct).
For example, having infinite streams data Stream a = Cons a (Stream a) and another applicative g, both Stream (g a) and g (Stream a) are applicatives.
But even though Stream is also a monad (join takes the diagonal of a 2-dimensional stream), its composition with another monad m won't be, neither Stream (m a) nor m (Stream a) will always be a monad.
Furthermore as we can see, they're both different from your MStream g (which is very close to ListT done right), therefore:
Ad 2.: Can all applicatives be constructed from some given set of primitives? Apparently not. One problem is constructing sum data types: If f and g are applicatives, Either (f a) (g a) won't be, as we don't know how to compose Right h <*> Left x.
Another construction primitive is taking a fixed point, as in your MStream example. Here we might attempt to generalize the construction by defining something like
newtype Fix1 f a = Fix1 { unFix1 :: f (Fix1 f) a }
instance (Functor (f (Fix1 f))) => Functor (Fix1 f) where
fmap f (Fix1 a) = Fix1 (fmap f a)
instance (Applicative (f (Fix1 f))) => Applicative (Fix1 f) where
pure k = Fix1 (pure k)
(Fix1 f) <*> (Fix1 x) = Fix1 (f <*> x)
(which requires not-so-nice UndecidableInstances) and then
data MStream' f g a = MStream (f (a, g a))
type MStream f = Fix1 (MStream' f)

A MonadTransControl instance for FreeT

Is it possible to implement a MonadTransControl instance for FreeT? I started with the following, but got stuck:
instance (Functor f) => MonadTransControl (FreeT f) where
newtype StT (FreeT f) r = FreeTStT r
liftWith unlift = lift $ unlift $ error "Stuck here"
restoreT inner = do
FreeTStT r <- lift inner
return r
If it is unimplementable, than why and is it possible to extend a specific free functor implementation somehow to make it implementable?
Disclaimer: turns out you need Traversable f constraint for MonadTransControl instance.
Warning: the instance in this answer does not obey all the laws of MonadTransControl
Pragmas and imports
{-# LANGUAGE TypeFamilies #-}
import qualified Data.Traversable as T
import Control.Monad
import Control.Monad.Trans
import Control.Monad.Trans.Control
import Control.Monad.Trans.Free
import qualified Control.Monad.Free as F
Free monadic state
As I said in comments, the proper "monadic state" of FreeT f should be Free f (the one from Control.Monad.Free):
instance T.Traversable f => MonadTransControl (FreeT f) where
newtype StT (FreeT f) a = StTFreeT { getStTFreeT :: F.Free f a }
Now the implementation of restoreT changes a bit:
restoreT inner = do
StTFreeT m <- lift inner
F.toFreeT m
liftWith implementation
Before we look at the implementation let's see what should the type of liftWith be:
liftWith :: Monad m => (Run (FreeT f) -> m a) -> FreeT f m a
And Run (FreeT f) is actually
forall n b. Monad n => FreeT f n b -> n (StTFreeT f b)
So the implementation would be like that:
liftWith unlift = lift $ unlift (liftM StTFreeT . pushFreeT)
The rest is simple:
pushFreeT :: (T.Traversable f, Monad m) => FreeT f m a -> m (F.Free f a)
pushFreeT m = do
f <- runFreeT m
case f of
Pure x -> return (return x)
Free y -> liftM wrap $ T.mapM pushFreeT y
Why Traversable?
As you can see the problem is with pushFreeT function: it uses T.mapM (which is traverse but with Monad constraint). Why do we need it there? If you look at the definition of FreeT you may notice that (NB: this is rough, I forget about Pure here):
FreeT f m a ~ m (f (m (f ... )))
And as a result of pushFreeT we need m (Free f a):
m (Free f a) ~ m (f (f (f ... )))
So we need to "push" all fs to the end and join all ms in the head. Thus we need an operation that lets us push a single f through single m and this is exactly what T.mapM pushFreeT gives us:
mapM :: (Monad m, Traversable t) => (a -> m b) -> t a -> m (t b)
mapM pushFreeT :: Traversable t => t (FreeT t m a) -> m (t (Free t a))
The laws
Every class instance usually come with laws. MonadTransControl is not an exception, so let's check if they hold for this instance:
liftWith . const . return = return
liftWith (const (m >>= f)) = liftWith (const m) >>= liftWith . const . f
These two laws obviously follow from laws for MonadTrans and the definition of liftWith.
liftWith (\run -> run t) >>= restoreT . return = t
Apparently, this law does not hold. This is because monad layers in t are collapsed when we pushFreeT. So the implemented liftWith merges effects in all layers of FreeT f m leaving us with the equivalent of m (Free f).

What would be the "distinct method" that Traversable has in addition to Foldable?

Foldable is a superclass of Traversable, similarly to how Functor is a superclass of Applicative and Monad.
Similar to the case of Monad, where it is possible to basically implement fmap as
liftM :: Monad m => (a->b) -> m a -> m b
liftM f q = return . f =<< q
we could also emulate foldMap as
foldLiftT :: (Traversable t, Monoid m) => (a -> m) -> t a -> m
foldLiftT f = fst . traverse (f >>> \x -> (x,x))
-- or: . sequenceA . fmap (f >>> \x -> (x, x))
using the Monoid m => (,) m monad. So the combination of superclass and methods bears in both cases a certain redundancy.
In case of monads, it can be argued that a “better” definition of the type class would be (I'll skip applicative / monoidal)
class (Functor m) => Monad m where
return :: a -> m a
join :: m (m a) -> m a
at least that's what's used in category theory. This definition does, without using the Functor superclass, not permit liftM, so it is without this redundancy.
Is a similar transformation possible for the Traversable class?
To clarify: what I'm after is a re-definition, let's call it,
class (Functor t, Foldable t) => Traversable t where
skim :: ???
such that we could make the actual Traverse methods top-level functions
sequenceA :: (Traversable t, Applicative f) => t (f a) -> f (t a)
but it would not be possible to make generically
instance (Traversable t) => Foldable t where
foldMap = ... skim ...
data T
instance Traversable T where
skim = ...
I'm not asking because I need this for something particular; it's a conceptual question so as to better understand the difference between Foldable and Traversable. Again much like Monad vs Functor: while >>= is much more convenient than join for everyday Haskell programming (because you usually need precisely this combination of fmap and join), the latter makes it simpler to grasp what a monad is about.
Foldable is to Functor as Traversable is to Monad, i.e. Foldable and Functor are superclasses of Monad and Traversable (modulo all the applicative/monad proposal noise).
Indeed, that's already in the code
instance Foldable f => Traversable f where
...
So, it's not clear what more there is to want. Foldable is characterized by toList :: Foldable f => f a -> [a] while Traversable depends ultimately on not only being able to abstract the content as a list like toList does, but also to be able to extract the shape
shape :: Functor f => f a -> f ()
shape = fmap (const ())
and then recombine them
combine :: Traversable f => f () -> [a] -> Maybe (f a)
combine f_ = evalStateT (traverse pop f_) where
pop :: StateT [a] Maybe a
pop = do x <- get
case x of
[] = empty
(a:as) = set as >> return a
which depends on traverse.
For more information on this property see this blog post by Russell O'Connor.
Super hand-wavy because it's late, but the extra power that Traversable has over Foldable is a way to reconstruct the original structure. For example, with lists:
module MyTraverse where
import Data.Foldable
import Data.Traversable
import Control.Applicative
import Data.Monoid
data ListRec f x = ListRec
{ el :: f (Endo [x])
}
instance Applicative f => Monoid (ListRec f x) where
mempty = ListRec (pure mempty)
mappend (ListRec l) (ListRec r) =
ListRec (mappend <$> l <*> r)
toM :: Functor f => f b -> ListRec f b
toM this = ListRec $ (Endo . (:)) <$> this
fromM :: Functor f => ListRec f b -> f [b]
fromM (ListRec l) = flip appEndo [] <$> l
myTraverse :: Applicative f => (a-> f b) -> [a] -> f [b]
myTraverse f xs = fromM $ foldMap (toM . f) xs
I think this myTraverse behaves the same as traverse, using only the classes Applicative, Foldable, and Monoid. You could re-write it to use foldr instead of foldMap if you wanted to get rid of Monoid.
lists are easy because they're a flat structure. However, I strongly suspect that you could use a Zipper to get the proper reconstruction function for any structure (since zippers are generically derivable, they should always exists).
But even with a zipper, you don't have any way of indicating that structure to the monoid/function. Notionally, it seems Traversable adds something like
class Traversed t where
type Path t :: *
annotate :: t a -> [(Path t, a)]
fromKeyed :: [(Path t, a)] -> t a
this seems to overlap heavily with Foldable, but I think that's inevitable when trying to associate the paths with their constituent values.

Is it possible to implement `(Applicative m) => Applicative (StateT s m)`?

I'm currently working on Data.Fresh and Control.Monad.Trans.Fresh, which resp. define an interface for generating fresh variables, and a monad transformer which implements this interface.
I initially thought it would be possible to implement the Applicative instance for my FreshT v m with the only requirement that Applicative m exists. However, I got stuck and it seemed like I need to require Monad m. Not trusting my Haskell-fu, I then turned to the transformers package, and was surprised by what I found in Control.Monad.Trans.State.Lazy and .Strict:
instance (Functor m, Monad m) => Applicative (StateT s m) where
pure = return
(<*>) = ap
So here is my question: is it possible to create an instance with equivalent semantics with the following instance head?
instance (Applicative m) => Applicative (StateT s m) where
Consider that you have two functions:
f :: s -> m (s, a -> b)
g :: s -> m (s, a)
And you want to create a function h = StateT f <*> StateF g
h :: s -> m (s, b)
From the above you have an s you can pass to f so you have:
f' :: m (s, a -> b)
g :: s -> m (s, a)
However to get s out of f' you need the Monad (whatever you'd do with applicative it would still be in form of m s so you would not be able to apply the value to g).
You can play with the definitions and use free monad but for the collapse of state you need join.
Weaker variant of an Applicative transformer
Although it isn't possible to define an applicative transformer for StateT, It's possible to define a weaker variant that works. Instead of having s -> m (a, s), where the state decides the next effect (therefore m must be a monad), we can use m (s -> (a, s)), or equivalently m (State s a).
import Control.Applicative
import Control.Monad
import Control.Monad.State
import Control.Monad.Trans
newtype StateTA s m a = StateTA (m (State s a))
This is strictly weaker than StateT. Every StateTA can be made into StateT (but not vice versa):
toStateTA :: Applicative m => StateTA s m a -> StateT s m a
toStateTA (StateTA k) = StateT $ \s -> flip runState s <$> k
Defining Functor and Applicative is just the matter of lifting operations of State into the underlying m:
instance (Functor m) => Functor (StateTA s m) where
fmap f (StateTA k) = StateTA $ liftM f <$> k
instance (Applicative m) => Applicative (StateTA s m) where
pure = StateTA . pure . return
(StateTA f) <*> (StateTA k) = StateTA $ ap <$> f <*> k
And we can define an applicative variant of lift:
lift :: (Applicative m) => m a -> StateTA s m a
lift = StateTA . fmap return
Update: Actually the above isn't necessary, as the composition of two applicative functors is always an applicative functor (unlike monads). Our StateTA is isomorphic to Compose m (State s), which is automatically Applicative:
instance (Applicative f, Applicative g) => Applicative (Compose f g) where
pure x = Compose (pure (pure x))
Compose f <*> Compose x = Compose ((<*>) <$> f <*> x)
Therefore we could write just
{-# LANGUAGE GeneralizedNewtypeDeriving #-}
import Control.Applicative
import Control.Monad.State
import Data.Functor.Compose
newtype StateTA s m a = StateTA (Compose m (State s) a)
deriving (Functor, Applicative)
Although, as noted in the previous answer, this instance cannot be defined in general, it is worth noting that, when f is Applicative and s is a Monoid, StateT s f is also Applicative, since it can be regarded as a composition of applicative functors:
StateT s f = Reader s `Compose` f `Compose` Writer s

Applicative Instance for (Monad m, Monoid o) => m o?

Sorry for the terrible title. I'm trying to make an instance of Applicative for a Monad wrapping a type that is a Monoid.
instance (Monad m, Monoid o) => Applicative (m o) where
pure x = return mempty
xm <*> ym = do
x <- xm
y <- ym
return $ x `mappend` y
This doesn't work; GCHi complains with:
Kind mis-match
The first argument of `Applicative' should have kind `* -> *',
but `m o' has kind `*'
In the instance declaration for `Applicative (m o)'
I realise that what I've written above may make no sense. Here is the context: I am trying to use the compos abstraction as described in the paper A pattern for almost compositional functions. Taking this tree (using the GADT version of compos; I've simplified it a lot):
data Tree :: * -> * where
Var :: String -> Expr
Abs :: [String] -> Expr -> Expr
App :: Expr -> [Expr] -> Expr
class Compos t where
compos :: Applicative f => (forall a. t a -> f (t a)) -> t c -> f (t c)
instance Compos Tree where
compos f t =
case t of
Abs ps e -> pure Abs <*> pure ps <*> f e
App e es -> pure App <*> f e <*> traverse f es
_ -> pure t
I'm going to write a lot of functions which descend the tree and return a list of say errors or a set of strings whilst also requiring state as it goes down (such as the binding environment), such as:
composFoldM :: (Compos t, Monad m, Monoid o) => (forall a. t a -> m o) -> t c -> m o
composFoldM f = ???
checkNames :: (Tree a) -> State (Set Name) [Error]
checkNames e =
case e of
Var n -> do
env <- get
-- check that n is in the current environment
return $ if Set.member n env then [] else [NameError n]
Abs ps e' -> do
env <- get
-- add the abstractions to the current environment
put $ insertManySet ps env
checkNames e'
_ -> composFoldM checkNames e
data Error = NameError Name
insertManySet xs s = Set.union s (Set.fromList xs)
I think these should all be able to be abstracted away by making composFoldM use compos for the (Monad m, Monoid o) => m o structure. So to use it with the GADT Applicative version of compos found on page 575/576 of the paper. I think I need to make an Applicative instance of this structure. How would I do this? Or am I going down completely the wrong path?
You want the Constant applicative from Data.Functor.Constant in the transformers package, which you can find here.
This Applicative has the following instance:
instance (Monoid a) => Applicative (Constant a) where
pure _ = Constant mempty
Constant x <*> Constant y = Constant (x `mappend` y)
You can then compose Constant with any other applicative using Compose from Data.Functor.Compose (also in the transformers package), which you can find here.
Compose has this Applicative instance:
instance (Applicative f, Applicative g) => Applicative (Compose f g) where
pure x = Compose (pure (pure x))
Compose f <*> Compose x = Compose ((<*>) <$> f <*> x)
You can then Compose your Constant applicative with any other Applicative (like State) to keep both some state and a running Monoid tally.
More generally, you should read the paper The Essence of the Iterator Pattern, which discusses these patterns in more detail.

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