/Runhaskell@Runhaskellbot {-# LANGUAGE ImplicitParams #-} foo :: Monad m => m () foo

Question

/Runhaskell@Runhaskellbot {-# LANGUAGE ImplicitParams #-} foo :: Monad m => m () foo

= let ?someData = 7 in kek
where
kek = do
xyz [?someData, (6 :: Int), 7, 8]

xyz :: Monad m => a -> m ()
xyz _ = pure ()

main = foo

#haskell #programming #russian

0

06.10.2023

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Danil Berestov · Answer 1

Main.hs:8:12: error: • Could not deduce: ?someData::Int arising from a use of implicit parameter ‘?someData’ from the context: Monad m bound by the type signature for: foo :: forall (m :: * -> *). Monad m => m () at Main.hs:4:1-22 • In the expression: ?someData In the first argument of ‘xyz’, namely ‘[?someData, (6 :: Int), 7, 8]’ In a stmt of a 'do' block: xyz [?someData, (6 :: Int), 7, 8] | 8 | xyz [?someData, (6 :: Int), 7, 8] | ^^^^^^^^^ Exit code: 1

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/Runhaskell@Runhaskellbot {-# LANGUAGE ImplicitParams #-} foo :: Monad m => m () foo

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