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

[Tony Morris]

FYI, the material you are using was stolen. Here is the original.

http://blog.tmorris.net/posts/20-intermediate-haskell-exercises/


On 23/10/15 22:42, Umair Saeed 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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