`Compose` allows you to join together `Flow a b` and `Flow b c`, regardless of what `b` is, even if it doesn't have a `Show` instance, so how can you possibly show it in the recursive call `runFlow f2`? Changing the definition of `Flow` to data Flow a b where Id :: Flow a a Compose :: Show b => Flow a b -> Flow b c -> Flow a c would allow you to write `runFlow` but then you can't define a `Category` instance, since `(.)` is not allowed to be constrained. Perhaps you want something like this: {-# LANGUAGE GADTs #-} import Control.Category import Prelude hiding (id, (.)) data Flow a b where Id :: Flow a a Compose :: Flow a b -> Flow b c -> Flow a c ShowId :: Show a => Flow a a instance Category Flow where id = Id (.) = flip Compose runFlow :: Flow a b -> a -> IO b runFlow f x = case f of Id -> pure x Compose f1 f2 -> runFlow f1 x >>= runFlow f2 ShowId -> print x >> pure x main :: IO () main = runFlow (ShowId >>> ShowId) () Tom On Fri, Jun 02, 2023 at 01:32:47PM +0000, Zoran Bošnjak wrote:

Dear haskell cafe members, I would appreciate a suggestion how to fix compile error on this simple test program (I am using ghc 9.0.2).

The idea is to have 'data Flow a b' unrestricted and create necessary constraints only when running/interpreting the flow. The problem is obviously an intermediate type 'b' in 'Compose', where the 'Show' instance is not deduced. I have a vague clue that some type families might be necessary to propagate Show constraint, or a type class with associated type family, but I don't know exactly how.

---

import Prelude hiding ((.), id) import Control.Category

data Flow a b where Id :: Flow a a Compose :: Flow a b -> Flow b c -> Flow a c

instance Category Flow where id = Id (.) = flip Compose

runFlow :: Show a => Flow a b -> a -> IO b runFlow f x = case f of Id -> print x >> pure x Compose f1 f2 -> runFlow f1 x >>= runFlow f2

main :: IO () main = runFlow (Id >>> Id) ()

---

The error is:

• Could not deduce (Show b1) arising from a use of ‘runFlow’ from the context: Show a bound by the type signature for: runFlow :: forall a b. Show a => Flow a b -> a -> IO b at a02.hs:13:1-42 Possible fix: add (Show b1) to the context of the data constructor ‘Compose’ • In the second argument of ‘(>>=)’, namely ‘runFlow f2’ In the expression: runFlow f1 x >>= runFlow f2 In a case alternative: Compose f1 f2 -> runFlow f1 x >>= runFlow f2 | 16 | Compose f1 f2 -> runFlow f1 x >>= runFlow f2 | ^^^^^^^^^^

On Fri, 2 Jun 2023, Zoran Bošnjak wrote:

import Prelude hiding ((.), id) import Control.Category

data Flow a b where Id :: Flow a a Compose :: Flow a b -> Flow b c -> Flow a c

type variable 'b' of Compose is not visible outside Flow. Thus you cannot get later in runFlow, but you have to embed the Show constraint in the Compose constructor like so: Compose :: (Show b) => Flow a b -> Flow b c -> Flow a c But this in turn means, you can compose only Flows where the interim type 'b' is an instance of Show. To avoid this you would need an additional type parameter to Flow with a constraint kind constructor or an existentially quantified type, that holds all the constraints you need for your current application. I think it should be like so: type family FlowConstraints constr a data FlowAny type instance FlowConstraints FlowAny a = () data FlowShow type instance FlowConstraints FlowShow a = (Show a) data Flow constr a b where Id :: Flow a a Compose :: (FlowConstraints constr b) => Flow a b -> Flow b c -> Flow a c or alternatively: data family FlowConstraints constr a data FlowAny data instance FlowConstraints FlowAby a = FlowAnyConstraint data FlowShow data instance FlowConstraints FlowShow a = (Show a) => FlowShowConstraint data Flow constr a b where Id :: Flow a a Compose :: (FlowConstraints constr b) => Flow a b -> Flow b c -> Flow a c In this case you have to match on FlowAnyConstraint or FlowShowConstraint in runFlow in order to get back the required constraints. In any case, the constraints must already be available at construction with Compose. Thus you will not be able to use Compose in a Category instance.