The problem is that in p you are using do notation for bind, which uses the Monad instance for bind (>>=) for ((->) String), because Parser is a type alias for (String -> [(a, String)].  But then you are using your own return which is not of the same type as Monad's return would be.  The way you have it now your return considers the loose `a` as the Monad parameter, but do notation considers it [(a, String)] instead.

You can either a) write your own bind that conforms to your type.

bind :: Parser a -> (a -> Parser b) -> Parser b
bind = undefined

p' :: Parser (Char, Char)
p' = item `bind` \x -> item `bind` \_ -> item `bind` \y -> Foo.return (x, y)

Or you can use the Monad return instead.  But the type is not what you expect.

p :: String -> ([(Char, String)], [(Char, String)])
p = do x <- item
       item
       y <- item
       Prelude.return (x,y)





On Mon, Nov 6, 2017 at 3:15 AM, Marcus Manning <iconsize@gmail.com> wrote:
type Parser a = String → [(a,String)]

item :: Parser Char 
item = λinp → case inp of 
                            [] → [] 
                            (x:xs) → [(x,xs)]
failure :: Parser a 
failure = λinp → []

return :: a → Parser a 
return v = λinp → [(v,inp)]

(+++) :: Parser a → Parser a → Parser a 
p +++ q = λinp → case p inp of 
                                 [] → q inp 
                                 [(v,out)] → [(v,out)]

parse :: Parser a → String → [(a,String)]
parse p inp = p inp

p :: Parser (Char,Char)
p = do x ← item 
           item 
           y ← item 
           return (x,y)

It is described in pages 189-216 in [1].


I assume the bind operator (==>) was overwritten by

(>>=) :: Parser a → (a → Parser b) → Parser b p 
>>= f = λinp → case parse p inp of 
                             [ ] → [ ] 
                             [ (v, out) ] → parse (f v) out

in order to manipulate the do expr to make the p function work, right?

2017-11-05 21:56 GMT+01:00 Tobias Brandt <to_br@uni-bremen.de>:

Hey,

can you show us your Parser definition? 

Cheers,
Tobias 

----- Nachricht von Marcus Manning <iconsize@gmail.com> ---------
     Datum: Sun, 5 Nov 2017 18:51:57 +0100
       Von: Marcus Manning <iconsize@gmail.com>
Antwort an: The Haskell-Beginners Mailing List - Discussion of primarily beginner-level topics related to Haskell <beginners@haskell.org>
   Betreff: [Haskell-beginners] Could not get parser ready
        An: beginners@haskell.org

Hello,

I follow the instructions of script [1] in order to set up a parser functionality. But I' get into problems at page 202 with the code:

p :: Parser (Char,Char)
p = do
             x ← item
             item
             y ← item
             return (x,y)


ghci and ghc throw errors:
Prelude> let p:: Parser (Char,Char); p = do {x <- item; item; y <- item; return (x,y)}

<interactive>:10:65: error:
    • Couldn't match type ‘[(Char, String)]’ with ‘Char’
      Expected type: String -> [((Char, Char), String)]
        Actual type: Parser ([(Char, String)], [(Char, String)])
    • In a stmt of a 'do' block: return (x, y)
      In the expression:
        do x <- item
           item
           y <- item
           return (x, y)
      In an equation for ‘p’:
          p = do x <- item
                 item
                 y <- item
                 ....
Did the semantics of do expr changed?

[1] https://userpages.uni-koblenz.de/~laemmel/paradigms1011/resources/pdf/haskell.pdf

Cheers,

iconfly.
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