Re: [Haskell-cafe] Partially applied type synonyms

Hello Tom, Type synonyms must be fully applied. You could try

newtype Foo (s :: * -> *) (x :: *) = MkFoo (Foo' (s x) x)

to make something that does not need to be fully applied -- but now you have to worry about the pesky MkFoo constructor. It's hard for me to suggest something else without understanding your use-case better. Sorry! Hope this helps, Richard

On Oct 7, 2021, at 10:15 PM, Ttt Mmm via Haskell-Cafe wrote:

I was surprised to find the below code doesn't typecheck even with -XLiberalTypeSynonyms. Am I missing something or is this really not possible?

Thanks, Tom

---

{-# LANGUAGE FlexibleInstances, KindSignatures, LiberalTypeSynonyms, StandaloneDeriving #-} -- This works: data Foo s x = Foo (s x) x deriving (Eq) -- This replacement doesn't: {- data Foo' sx x = Foo' sx x deriving (Eq) type Foo (s :: * -> *) (x :: *) = Foo' (s x) x -} data Bar (m :: * -> *) = Bar (m Int)

-- Neither of these typecheck: x :: Bar (Foo Maybe) x = undefined deriving instance Eq (Bar (Foo Maybe))

_______________________________________________ Haskell-Cafe mailing list To (un)subscribe, modify options or view archives go to: http://mail.haskell.org/cgi-bin/mailman/listinfo/haskell-cafe Only members subscribed via the mailman list are allowed to post.