glynn.clements at virgin.net
Sun Dec 28 08:27:19 EST 2003
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 <glynn.clements at virgin.net>
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