Hi all, I've been reading this article: http://perl.plover.com/lambda/tpj.html in which the author talks about the lambda calculus and uses examples in Perl (here's a link directly to the code: http://perl.plover.com/lambda/lambda-brief.pl) I'm a total newbie with respect to the lambda calculus. fA = \x -> 'A' fB = \x -> 'B' x -=- y = (show $ x id) == (show $ y id) I tried (naively) to port these examples to Haskell to play a bit with them: fTRUE = \a -> \b -> a fFALSE = \a -> \b -> b fIF = \b -> \x -> \y -> b x y fPAIR = \a -> \b -> \f -> f a b fFIRST = \p -> p fTRUE fSECOND = \p -> p fFALSE fZERO = fPAIR fTRUE fTRUE fSUCC = \x -> fPAIR fFALSE x fIS_ZERO = \x -> fFIRST x fPRED = \x -> fSECOND x fONE = fSUCC fZERO fTWO = fSUCC fONE fADD = \m -> (\n -> fIF (fIS_ZERO m) n (fADD (fPRED m) (fSUCC n))) but I couldn't get fADD to compile: Occurs check: cannot construct the infinite type: t = (t1 -> t1 -> t1) -> t Probable cause: `fADD' is applied to too many arguments In the third argument of `fIF', namely `(fADD (fPRED m) (fSUCC n))' In the expression: fIF (fIS_ZERO m) n (fADD (fPRED m) (fSUCC n)) I think its because in these Perl examples all functions are treated as being of the same type (number or type of args doesn't matter), but this is not the case in Haskell. Is there any way to create code similar to this in Haskell? Thanks, Patrick -- ===================== Patrick LeBoutillier Rosemère, Québec, Canada