Hi, You can only use do notation if you actually create an instance of Monad, which for Parser you haven't done. To continue as is, replace the first line with:
import Prelude hiding (return, fail, (>>=)) and the p function with p = item >>= \x -> item >>= \_ -> item >>= \y -> return (x, y) I've basically de-sugared the do-notation you wrote and hid the >>= from Prelude so that the one you declared locally is used. Michael On Tue, Mar 16, 2010 at 9:09 PM, 国平张 <zhangguop...@gmail.com> wrote: > Hi, > > I am a beginner for haskell. I was stuck with a sample of "programming > in haskell". Following is my code: > --------------------------------------------------------------------- > import Prelude hiding (return, fail) > > type Parser a = (String->[(a,String)]) > > return :: a -> Parser a > return v = (\inp->[(v,inp)]) > > item :: Parser Char > item = \inp -> case inp of > [] -> [] > (x:xs) -> [(x,xs)] > failure :: Parser a > failure = \inp -> [] > > parse :: Parser a->(String->[(a,String)]) > parse p inp = p inp > > (>>=) :: Parser a -> (a -> Parser b) -> Parser b > p >>= f = (\inp -> case parse p inp of > [] -> [] > [(v,out)]->parse (f v) out) > > p :: Parser (Char,Char) > p = do x <- item > item > y <- item > return (x,y) > --------------------------------------------------------------------- > > But it cannot be loadded by Hug, saying: > > Couldn't match expected type `Char' > against inferred type `[(Char, String)]' > Expected type: [((Char, Char), String)] > Inferred type: [(([(Char, String)], [(Char, String)]), String)] > In the expression: return (x, y) > In the expression: > do x <- item > item > y <- item > return (x, y) > > ------------------------------------------------------------------- > > I googled and tried a few days still cannot get it compiled, can > someone do me a favor to point out what's wrong with it :-) ? > _______________________________________________ > Haskell-Cafe mailing list > Haskell-Cafe@haskell.org > http://www.haskell.org/mailman/listinfo/haskell-cafe >
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