Hello all,
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