Re: [Haskell-beginners] Beginners Digest, Vol 106, Issue 7

That depends on what package you are using to parse. If you are using parsec, you can use the string function from Text.Parsec.Char. If you are using some other package, it probably has a different name for it. On Tue, Apr 18, 2017 at 8:39 AM, Andrey Klaus wrote:

Hello everybody,

A small question. ----- packageP = do literal “package" -----

what is the "literal" in this code? My problem is

$ ghc ParserTest.hs [1 of 1] Compiling ParserTest ( ParserTest.hs, ParserTest.o )

ParserTest.hs:11:5: Not in scope: ‘literal’

$ ghc --version The Glorious Glasgow Haskell Compilation System, version 7.10.3

Is this because I use old version of software?

Thanks, Andrey

2017-04-14 21:58 GMT+03:00 :

...

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Today's Topics:

1. Parsing (mike h) 2. Re: Parsing (David McBride) 3. Re: Parsing (Francesco Ariis) 4. Re: Parsing (mike h) 5. Re: Parsing (mike h)

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Message: 1 Date: Fri, 14 Apr 2017 19:02:37 +0100 From: mike h To: The Haskell-Beginners Mailing List - Discussion of primarily beginner-level topics related to Haskell Subject: [Haskell-beginners] Parsing Message-ID: <2C66C9DC-30AF-41C5-B9AF-0D1DA19E0A2C@yahoo.co.uk> Content-Type: text/plain; charset=utf-8

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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Message: 2 Date: Fri, 14 Apr 2017 14:17:42 -0400 From: David McBride To: The Haskell-Beginners Mailing List - Discussion of primarily beginner-level topics related to Haskell Subject: Re: [Haskell-beginners] Parsing Message-ID:

Content-Type: text/plain; charset=UTF-8

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

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

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Message: 3 Date: Fri, 14 Apr 2017 20:35:32 +0200 From: Francesco Ariis To: beginners@haskell.org Subject: Re: [Haskell-beginners] Parsing Message-ID: <20170414183532.GA4376@casa.casa> Content-Type: text/plain; charset=utf-8

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...

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Message: 4 Date: Fri, 14 Apr 2017 20:12:14 +0100 From: mike h To: The Haskell-Beginners Mailing List - Discussion of primarily beginner-level topics related to Haskell Subject: Re: [Haskell-beginners] Parsing Message-ID: Content-Type: text/plain; charset=utf-8

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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Message: 5 Date: Fri, 14 Apr 2017 20:19:40 +0100 From: mike h To: The Haskell-Beginners Mailing List - Discussion of primarily beginner-level topics related to Haskell Subject: Re: [Haskell-beginners] Parsing Message-ID: Content-Type: text/plain; charset="utf-8"

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