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. I hope the combination of keywords in my
subject line can serve to direct someone else with this problem to your
answer!
Todd Wilson
On Tue, Nov 24, 2020 at 2:37 PM Adam Gundry
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
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