Lee Dixon wrote:
I've run into a small problem whilst doing some manual type checking, to see if I could match the results given by hugs
prelude> :t foldr filter foldr filter :: [a] -> [a -> Bool] -> [a] -- This was fine and was the same as my answer, so I tested it with prelude> foldr filter [1,2,3,4] [even,odd] -- and the answer was indeed an empty list
-- However I got stuck when attempting to derive the result of map (foldr filter) -- Hugs says that: prelude> :t map (foldr filter) map (foldr filter) :: [[a]] -> [[a -> Bool] -> [a]]
Two main questions: 1/ How does hugs derive this answer?
filter :: (a -> Bool) -> [a] -> [a] foldr :: (a -> b -> b) -> b -> [a] -> b map :: (a -> b) -> [a] -> [b] Passing filter as the argument to foldr instantiates a as (a -> Bool) and b as [a], giving: foldr filter :: [a] -> [a -> Bool] -> [a] Passing (foldr filter) as the argument to map instantiates a as [a] and b as ([a -> Bool] -> [a]), giving: map (foldr filter) :: [[a]] -> [[a -> Bool] -> [a]]
2/ What input can I give so that it yields a correct result? I've tried giving it a list of lists but it fails...
The return value is a list of functions, and functions aren't
instances of Show, so it can't print the result.
Maybe you didn't mean (map (foldr filter))? The argument to map is
usually a function of one argument (i.e. a function whose result
*isn't* a function).
What are you ultimately trying to achieve?
--
Glynn Clements