[Haskell-beginners] Why can't I return a partially applied function in my example?

[Andres Loeh]

Hi.

You are right, that 'f (g x)' has type '(t -> b)'. However, you're not
returning 'f (g x)'. You're returning '\x -> f (g x)', and that has
type 't -> (t -> b)', because 'x' is of type 't'. So you're returning
a 't -> (t -> b)' where a 't -> b' is expected. Since the 't -> ...'
of both types matches, GHC complains only about the mismatch of the
result types.

Cheers,
  Andres

On Fri, Oct 23, 2015 at 2:42 PM, Umair Saeed <umairsd at gmail.com> wrote:
> Hello all,
> I'm learning Haskell, and started to go through a set of intermediate
> exercises
> (https://www.fpcomplete.com/user/DanBurton/20-intermediate-exercises). I am
> a bit puzzled about one of the exercises, and hope someone can help me
> understand why one of my solutions doesn't work.
>
> We have a typeclass, Misty (only the relevant banana function shown) as:
>
>
> class Misty m where
>     banana :: (a -> m b) -> m a -> m b
>
>
> The exercise asks to implement this typeclass for the type ‘((->) t)’. I
> started off by filling in the relevant types, and I get:
>
> banana :: (a -> ((->) t b) ) -> ((->) t a) -> ((->) t b)
> banana :: (a -> (t -> b)) -> (t -> a) -> (t -> b)
>
> Based on this, I decided to implement banana as:
>
> banana f g = (\x -> f (g x))
>
> Here is my thought process:
> - The type of f is ‘(a -> t -> b)’, and the type of g is ‘(t -> a)’
> - g converts an argument of type ‘t’ into a result of type ‘a’.
> - I then pass the result of ‘(g x)’ (which is of type ‘a’) as an argument to
> ‘f’.
> - At this point, ‘f’ would be partially applied, and I *expect* to get a
> result of type ‘(t -> b)’
>
>
> However, when I try to build my solution, I get the following error (code is
> in a file called intermediate-help.hs):
>
> [1 of 1] Compiling Main             ( intermediate-help.hs, interpreted )
>
> intermediate-help.hs:7:25:
>     Couldn't match expected type ‘b’ with actual type ‘t -> b’
>       ‘b’ is a rigid type variable bound by
>           the type signature for
>             banana :: (a -> t -> b) -> (t -> a) -> t -> b
>           at intermediate-help.hs:7:5
>     Relevant bindings include
>       x :: t (bound at intermediate-help.hs:7:20)
>       g :: t -> a (bound at intermediate-help.hs:7:14)
>       f :: a -> t -> b (bound at intermediate-help.hs:7:12)
>       banana :: (a -> t -> b) -> (t -> a) -> t -> b
>         (bound at intermediate-help.hs:7:5)
>     In the expression: f (g x)
>     In the expression: (\ x -> f (g x))
> Failed, modules loaded: none.
>
>
>
> So here's my confusion: The compiler is complaining that it cannot match
> expected type ‘b’ with actual type ‘t -> b’. However, as I reasoned above,
> when I wrote this code, I expected to get type ‘t -> b’. Clearly, my thought
> process has a hole, and I need help/advice from more experienced Haskellers
> to identify what I am missing.
>
> Thank you for any help,
> ~Umair
>
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