Re: [Haskell-cafe] stuck with a sample of "programming in haskell"

17 Mar 2010


      Am Mittwoch 17 März 2010 16:35:08 schrieb 国平张:
...
Thanks very much. It works!
I just wonder if you can help me to define a Monad to make "do" notion
works :-) ?
To make an instance of Monad, you must create a new datatype, for example

module Parse where

newtype Parser a = P { parse :: (String -> [(a,String)]) }

instance Monad Parser where
    return v = P (\s -> [(v,s)])
    p >>= f = P (\s -> case parse p s of
                        []    -> []
                        [(v,str)] -> parse (f v) str)
    fail _ = P (\_ -> [])
...
I know it is bothering, but I just ever tried to define a Monad,
failed either. What I did to define a Monad was:
instance Monad Parser where
   return v = (\inp->[(v,inp)])
   f >>= g =  = (\inp -> case parse p inp of
                                    [] -> []
                                    [(v,out)]->parse (f v) out)
But it did not compile :-(.
Best Regards,
Guo-ping
2010/3/17 Michael Snoyman :
...
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  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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Re: [Haskell-cafe] stuck with a sample of "programming in haskell"

Daniel Fischer