Hi Dmitri,
your f1 function has 3 arguments, f, g and x.
you pass f as the first argument to mapMaybe, so it naturally must have type (a
-> b).
you pass the result of (g x) to the second argument of mapMaybe, so (g x) must
have type Maybe a. This means g must have the type (t -> Maybe a) where t is the
type of x.
This gives f1 :: (a -> b) -> (t -> Maybe a) -> t -> Maybe b
you are going to be passing in something with type (c -> Maybe d) as the first
argument to f1. (I used different type variables to reduce confusion)
This constraint gives f1 the following type
f1 :: (c -> Maybe d) -> (t -> Maybe c) -> t -> Maybe (Maybe d)
substituting different type variable names gives
f1 :: (b -> Maybe c) -> (a -> Maybe b) -> a -> Maybe (Maybe c)
So you are very close to finishing... :)
Hope this helps.
DeeJay
Dmitri O.Kondratiev wrote:
I am trying to solve a problem from "The Craft of Functional
Programming" book:
14.38 ... define the function:
data Maybe a = Nothing | Just a
composeMaybe :: (a -> Maybe b) -> (b -> Maybe c) -> (a -> Maybe c)
using functions:
squashMaybe :: Maybe (Maybe a) -> Maybe a
squashMaybe (Just (Just x)) = Just x
squashMaybe _ = Nothing
mapMaybe :: (a -> b) -> Maybe a -> Maybe b
mapMaybe f Nothing = Nothing
mapMaybe f (Just x) = Just (f x)
As a first step to the solution I defined auxilary function:
f1 f g x = mapMaybe f (g x)
GHCi gives the following type for this function:
f1 :: (a -> b) -> (t -> Maybe a) -> t -> Maybe b
^^^
Q: I don't quite understand this signature. I would expect this
instead (by mapMaybe definition):
f1 :: (a -> b) -> (t -> Maybe a) -> Maybe b
From where does the second 't' come from? What are the arguments and
what f1 returns in this case?
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