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
<zhangguoping@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 :-) ?
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