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
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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