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

6 Nov 2017

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

with

bind:: Parser a -> (a -> Parser b) -> Parser b
bind p f = \ inp -> case parse p inp of
                                [] -> []
                                [(v,out)] -> parse (f v) out

works like a charm.

Another alternatiev would to change the Parser definition to the one given
in [1].

Thanks.

Cheers,

iconfly

[1] http://dev.stephendiehl.com/fun/002_parsers.html

2017-11-06 14:29 GMT+01:00 David McBride :
...
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  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].
[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 :
...
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/reso
urces/pdf/haskell.pdf
Cheers,
iconfly.
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