Re: [Haskell-beginners] Parsing

Hi Francesco, Yes, I think you are right with "Are you sure you are not wanting [String] instead of String?” I could use Parsec but I’m building up a parser library from first principles i.e. newtype Parser a = P (String -> [(a,String)]) parse :: Parser a -> String -> [(a,String)] parse (P p) = p and so on…. It’s just an exercise to see how far I can get. And its good fun. So maybe I need add another combinator or to what I already have. Thanks Mike

On 14 Apr 2017, at 19:35, Francesco Ariis wrote:

On Fri, Apr 14, 2017 at 07:02:37PM +0100, 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

Hello Mike,

I am not really sure what you are doing here? You are parsing a dot separated list (like.this.one) but at the end you are concatenating all together, why? Are you sure you are not wanting [String] instead of String?

If so, Parsec comes with some handy parser combinators [1], maybe one of them could fit your bill:

-- should work packageP = literal "package" *> Pkg <$> sepEndBy1 identifier (char '.')

[1] https://hackage.haskell.org/package/parsec-3.1.11/docs/Text-Parsec-Combinato... https://hackage.haskell.org/package/parsec-3.1.11/docs/Text-Parsec-Combinato... _______________________________________________ Beginners mailing list Beginners@haskell.org mailto:Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners