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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