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 [EMAIL PROTECTED] _______________________________________________ Haskell mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell
