Re: [Haskell-beginners] Could not get parser ready

6 Nov 2017

      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/haskel...

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 :
...
Hey,
can you show us your Parser definition?
Cheers,
Tobias
----- Nachricht von Marcus Manning  ---------
     Datum: Sun, 5 Nov 2017 18:51:57 +0100
       Von: Marcus Manning 
Antwort an: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell 
   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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