[Haskell-cafe] could not deduce Show compile error

2 Jun 2023

      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
   |                                       ^^^^^^^^^^

kind regards,
Zoran

[Haskell-cafe] could not deduce Show compile error

Zoran Bošnjak