[Haskell-beginners] Unable to sequence using 'do' operator

[David McBride]

Your indentation is not correct.  Remember that haskell is whitespace
sensitive.

firstAndThird :: Parser (Char, Char)
firstAndThird = do
  x <- item
  item
  y <- item
  return (x,y)

should work.


On Mon, Nov 17, 2014 at 10:20 AM, Rohit Sharma <rohits79 at gmail.com> wrote:

> type Parser a = String -> [(a, String)]
>
> Hi,
>
> I am learning haskell and wondering why my definition of firstAndThird
> does not work with the do operator, however when i try to use the same
> using bind (firstandThird2) it works as expected. I basically want to
> return a tuple of first and third character of a string. Can someone please
> correct me what i might be overseeing?
>
> I have pasted the same code on codetidy in case if the email lose
> formatting.
> http://codetidy.com/5726/
>
> Many Thanks,
> Rohit
>
> zero :: Parser a
> zero = \inp -> []
>
> result :: a -> Parser a
> result x = \inp -> [(x, inp)]
>
> item :: Parser Char
> item = \inp -> case inp of
>                    [] -> []
>                    (x:xs) -> [(x,xs)]
>
> bind :: Parser a -> (a -> Parser b) -> Parser b
> p `bind` f = \inp -> concat [ ((f x) inp') | (x, inp') <- p inp]
>
> sat :: (Char -> Bool) -> Parser Char
> sat predicate = item `bind` (\x -> if predicate x then result x else zero )
>
> lower :: Parser Char
> lower = sat (\x -> 'a' <= x && x <= 'z')
>
> firstAndThird :: Parser (Char, Char)
> firstAndThird = do x <- item
>                         item
>                    y <- item
>                    return (x,y)
>
> firstAndThird2 :: Parser (Char, Char)
> firstAndThird2 = item `bind` \x ->
>       item `bind` \y ->
>       item  `bind` \z ->
>       result (x,z)
>
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