[Haskell-cafe] Newbie Q: composeMaybe :: (a -> Maybe b) -> (b -> Maybe c) -> (a -> Maybe c)

[DeeJay-G615]

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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