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. _______________________________________________ Beginners mailing list Beginners@haskell.orghttp://mail.haskell.org/cgi-bin/ mailman/listinfo/beginners

----- Ende der Nachricht von Marcus Manning -----

_______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners

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. _______________________________________________ Beginners mailing list Beginners@haskell.orghttp://mail.haskell.org/cgi-bin/mailman /listinfo/beginners

----- Ende der Nachricht von Marcus Manning -----

_______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners

_______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners

_______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners