Re: [Haskell-beginners] Parsing

Duh! I did have that - well clearly not exactly! What I’d done a couple of hours ago had the parens in the wrong place Earlier i had packageP' = literal "package" >> Pkg <$> (:) <$> identifier <*> many ((:) <$> char '.' <*> identifier) i.e. the first <$> (:) what I have now is (thanks David) packageP' = literal "package" >> Pkg <$> ((:) <$> identifier <*> many ((:) <$> char '.' <*> identifier)) :) Looking at it now what I first had is blindingly obviously wrong! Haskell often makes me feel stupid and makes me work for my code, thats why I love it! Cheers Mike

On 14 Apr 2017, at 20:27, David McBride wrote:

Working it out for myself, it should be something like:

blah :: forall f. Applicative f => f PackageDec blah = package *> Pkg . mconcat <$> ((:) <$> identifier <*> restOfIdentifiers) where restOfIdentifiers :: Applicative f => f [String] restOfIdentifiers = many ((:) <$> char '.' <*> identifier)

On Fri, Apr 14, 2017 at 3:19 PM, mike h wrote:

...

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

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