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
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