Re: [Haskell-beginners] Parsing

Hi David, Thanks but I tried something like that before I posted. I’ll try again maybe I mistyped. Mike

On 14 Apr 2017, at 19:17, David McBride wrote:

Try breaking it up into pieces. There a literal "package" which is dropped. There is a first identifier, then there are the rest of the identifiers (a list), then those two things are combined somehow (with :).

literal "package" *> (:) <$> identifier <*> restOfIdentifiers where restOfIdentifiers :: Applicative f => f [String] restOfIdentifiers = many ((:) <$> char '.' <*> identifier

I have not tested this code, but it should be close to what you are looking for.

On Fri, Apr 14, 2017 at 2:02 PM, mike h wrote:

...

I have data PackageDec = Pkg String deriving Show

and a parser for it

packageP :: Parser PackageDec packageP = do literal “package" x <- identifier xs <- many ((:) <$> char '.' <*> identifier) return $ Pkg . concat $ (x:xs)

so I’m parsing for this sort of string “package some.sort.of.name”

and I’m trying to rewrite the packageP parser in applicative style. As a not quite correct start I have

packageP' :: Parser PackageDec packageP' = literal "package" >> Pkg . concat <$> many ((:) <$> char '.' <*> identifier)

but I can’t see how to get the ‘first’ identifier into this sequence - i.e. the bit that corresponds to x <- identifier in the monadic version.

in ghci λ-> :t many ((:) <$> char '.' <*> identifier) many ((:) <$> char '.' <*> identifier) :: Parser [[Char]]

so I think that somehow I need to get the ‘first’ identifier into a list just after Pkg . concat so that the whole list gets flattened and everybody is happy!

Any help appreciated.

Thanks Mike

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