Hey DeeJay, Thanks for detailed answer, it really helps as it shows the way I need to follow! It also helped me to realize that my question why f1 f g x = mapMaybe f (g x) has this type: f1 :: (a -> b) -> (t -> Maybe a) -> t -> Maybe b and not that (type which I expected): f1 :: (a -> b) -> (t -> Maybe a) -> Maybe b
was actually very stupid of me, as I just blindly missed that f1 *also has* a third parameter 'x' of type 't' !!! (Perhaps that happened to me because I was too much absorbed by the grave nature of the problem of composing functions returning 'Maybe' :) Thanks again! Dima On 11/8/06, DeeJay-G615 <[EMAIL PROTECTED]> wrote:
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? > _______________________________________________ > Haskell-Cafe mailing list > Haskell-Cafe@haskell.org > http://www.haskell.org/mailman/listinfo/haskell-cafe > > >
-- Dmitri O Kondratiev, [EMAIL PROTECTED] http://www.geocities.com/dkondr/ _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
