On Tue, 24 Nov 2020, Viktor Dukhovni wrote:
On Tue, Nov 24, 2020 at 02:58:33PM -0800, Todd Wilson wrote:
I see, thanks! And I guess that if the type variable c in your type of g were changed to b, it will still work, but now g would have a monomorphic instead of polymorphic type.
Yes, exactly. I was not going to post my otherwise redundant answer, but since it in fact makes `g` monomorphic, here it is:
The simplest thing (which you probably already know) is to not give `g` a type signature, and let GHC infer the type. But since you asked, one solution is:
{-# LANGUAGE ScopedTypeVariables #-}
f :: forall a b. a -> [b] -> [(a, b)] f a bs = g bs where g :: [b] -> [(a, b)] g [] = [] g (x:xs) = (a, x) : g xs
An alternative would be to let g be polymorphic but add 'a' as parameter: f a bs = g a bs where g :: a -> [b] -> [(a, b)] g = map ((,) a)