Dear Todd, On 24/11/2020 22:26, Todd Wilson wrote:
This has got to be a trivial question, but I don't see a solution, nor could I find anything through searching: If I want to write a type declaration for the helper function g here, what do I write?
f :: a -> [b] -> [(a,b)] f a bs = g bs where g :: ???? g [] = [] g (x:xs) = (a,x) : g xs
This has a disappointingly not-entirely-trivial answer: {-# LANGUAGE ExplicitForAll, ScopedTypeVariables #-} f :: forall a b . a -> [b] -> [(a,b)] f a bs = g bs where g :: [c] -> [(a, c)] g [] = [] g (x:xs) = (a,x) : g xs We need the type variable `a` in the type of `g` to be the same as the one bound in the type of `f`, but Haskell2010 doesn't have a way to express this. The ScopedTypeVariables extension, when used with an explicit forall, makes the type variables scope over type signatures in the body of the function. Hope this helps, Adam -- Adam Gundry, Haskell Consultant Well-Typed LLP, https://www.well-typed.com/ Registered in England & Wales, OC335890 118 Wymering Mansions, Wymering Road, London W9 2NF, England