Question

As part of my Haskell learning process, I like to explicitly type out the type declarations for functions. I would like to be able to do so for functions defined in a where clause, but I don’t know how to specify, that a type variable in a where clause should denote the same type as some type variable in the outer type declaration. For instance, the following code:

foo :: (a -> a) -> a -> a
foo f arg = bar arg
  where
    bar :: a -> a
    bar a = f a

yields this error:

src\Test.hs:7:14:
    Couldn't match expected type `a' against inferred type `a1'
      `a' is a rigid type variable bound by
          the type signature for `foo' at src\Test.hs:3:8
      `a1' is a rigid type variable bound by
           the type signature for `bar' at src\Test.hs:6:11
    In the first argument of `f', namely `a'
    In the expression: f a
    In the definition of `bar': bar a = f a

How can I express that the first argument to bar should be of the same type as the second argument to foo, so that I can apply f to it?

Thanks.

Answer 1

I think you can do this in general with ScopedTypeVariables which GHC supports. This certainly compiles:

{-# LANGUAGE ScopedTypeVariables #-}
foo :: forall a. (a -> a) -> a -> a
foo f arg = bar arg
  where
    bar :: a -> a
    bar a = f a

Note the "forall a."