I've found out what was wrong. I should have written: perms (a:as) = concatMap (\b -> map ((:) (fst b)) (perms (snd b))) (del (a:as)) but I still don't understand why it had the error message it did. Ie, how did it infer the type of my lambda function to be "([a],[a]) -> [[a]]"? Cheers, Mark. Mark Phillips wrote:
Hi,
I have recently started learning Haskell and, in writing a HUGS module to generate permutations, have been told I have an error but I don't understand why.
The module is:
module Arrange where -- -- perms :: [a] -> [[a]] perms [] = [[]] perms (a:as) = concatMap (\b -> fst b:perms (snd b)) (del (a:as))
del :: [a] -> [(a,[a])] del [] = [] del (a:as) = (a,as):(map (\b -> (fst b,a:(snd b))) (del as))
and it comes back with error message:
Type checking ERROR Arrange.hs:6 - Type error in application *** Expression : concatMap (\b -> fst b : perms (snd b)) (del (a : as)) *** Term : \b -> fst b : perms (snd b) *** Type : ([a],[a]) -> [[a]] *** Does not match : (a,[a]) -> [[a]] *** Because : unification would give infinite type
But why does it say that the term "\b -> fst b : perms (snd b)" has type "([a],[a]) -> [[a]]"?
perms requires a type "[a]" as input so "snd b" should be of type "[a]" but "fst b" should be allowed to be anything.
What's going on?
Any help would be much appreciated.
Thanks,
Mark.
-- Dr Mark H Phillips Research Analyst (Mathematician) AUSTRICS - Smarter Scheduling Solutions - www.austrics.com Level 2, 50 Pirie Street, Adelaide SA 5000, Australia Phone +61 8 8226 9850 Fax +61 8 8231 4821 Email mark@austrics.com.au