Haskell: How to write a type of function from a specific type to any type? - haskell

Question

In Scala, I could write the following trait:
trait Consumer[A] {
def apply(a: A): Unit
}
And scala would convert whatever I want to Unit, i.e., it would discard the type. Equivalently, I could have said that apply returns Any and ignore the result.
However, in Haskell, if I defined the type as type Consumer = a -> IO (), I wouldn't be able to pass an Int -> IO Int function, as Int isn't ().
There are two ways I know of solving this issue, but none are satisfactory:
Use Data.Functor.void at the call site to manual change IO a to IO (). This is annoying as an API user.
define type Consumer a b = a -> IO b, but then every time I would want to use Consumer in a signature, I would have to carry the useless type b.
Is there any way to define the Consumer type as a function from a to "IO Any"? As far as I know, Haskell does not support something like exists x. a -> IO x.
Using forall results in the opposite of what I want, e.g.,
type Consumer = forall b. a -> IO b
foo :: Int -> IO Int
foo = undefined
bar :: Consumer Int
bar = foo
results in the error:
• Couldn't match type ‘b’ with ‘Int’
‘b’ is a rigid type variable bound by
the type signature for:
bar :: Consumer Int
Expected type: Int -> IO b
Actual type: Int -> IO Int
• In the expression: foo
In an equation for ‘bar’: bar = foo
• Relevant bindings include
bar :: Int -> IO b
Note that I specifically want Consumer to a be type alias, and not a data constructor, as is described here: Haskell function returning existential type. I wouldn't mind if Consumer were a class if anyone knows how to make that work.

Answer 1

To get an existentially-quantified type in Haskell, you need to write down a data declaration (as opposed to a newtype declaration or a type alias declaration, like you used.).
Here's a Consumer type that fits your purposes:
{-# LANGUAGE ExistentialQuantification #-}
data Consumer input = forall output. Consumer { runDiscardingOutput :: input -> IO output }
And, analogously, here is what your example would look like with the new type:
f :: Int -> IO Int
f = undefined
g :: Consumer Int
g = Consumer f
This doesn't really avoid your concerns about client code needing an extra call, though. (I mean, this is no better than exporting a consumer = Data.Functor.void binding from your library.) Also, it complicates how clients will be able to use a consumer, too:
consumer :: Consumer Int
consumer = Consumer (\x -> return [x])
{- This doesn't typecheck -}
main1 :: IO ()
main1 = runIgnoringOutput consumer 4
{- This doesn't typecheck (!!!) -}
main2 :: IO ()
main2 = void (runIgnoringOutput consumer 4)
{- Only this typechecks :( -}
main3 :: IO ()
main3 =
case consumer of
Consumer f -> Data.Functor.void (f 4)
So it would probably make sense to have a apply function in your library that did the dirty work, just as there was an apply function in the Scala library.
apply :: Consumer a -> a -> IO ()
apply (Consumer f) x = void (f x)

Answer 2

I wouldn't mind if Consumer were a class if anyone knows how to make that work.
You can simulate existential types for classes with an associated type family.
But Haskell doesn't allow ambiguous types in classes without using something like a GADT existential wrapper, so you would still have the type information there somewhere.
{-# LANGUAGE TypeFamilies, MultiParamTypeClasses #-}
class Consumer c a where
type Output c
consume :: c -> a -> IO (Output c)
c is necessary here to allow for the reconstruction of the type of Output c, so it is not strictly an existential. But you can now write
{-# LANGUAGE FlexibleInstances, InstanceSigs #-}
instance Consumer (a -> IO b) a where
type Output (a -> IO b) = b
consume :: (a -> IO b) -> a -> IO b
consume = id
This may not fit your use case, because there will not be a type signature that can express Consumer a in a truly existential way. But it is possible to write
... :: (Consumer c a) => c -> ...
(You could also make use of FunctionalDependencies here to clarify the class somewhat.)