[Haskell-beginners] Could not get parser ready

[Marcus Manning]

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].

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

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 at uni-bremen.de>:

> Hey,
>
> can you show us your Parser definition?
>
> Cheers,
> Tobias
>
> ----- Nachricht von Marcus Manning <iconsize at gmail.com> ---------
>      Datum: Sun, 5 Nov 2017 18:51:57 +0100
>        Von: Marcus Manning <iconsize at gmail.com>
> Antwort an: The Haskell-Beginners Mailing List - Discussion of primarily
> beginner-level topics related to Haskell <beginners at haskell.org>
>    Betreff: [Haskell-beginners] Could not get parser ready
>         An: beginners at 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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> Beginners at haskell.orghttp://mail.haskell.org/cgi-bin/
> mailman/listinfo/beginners
>
>
>
>
> ----- Ende der Nachricht von Marcus Manning <iconsize at gmail.com> -----
>
>
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