Can anyone suggest a good Haskell IDE for windows?
From: Beginners <beginners-bounces@haskell.org> on behalf of David McBride <toad3k@gmail.com>
Sent: Tuesday, April 18, 2017 4:22 PMSubject: Re: [Haskell-beginners] Beginners Digest, Vol 106, Issue 7
To: The Haskell-Beginners Mailing List - Discussion of primarily beginner-level topics related to Haskell
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 <deepmindster@gmail.com> 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 <beginners-request@haskell.org>:
>>
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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)
>>
>>
>> ------------------------------------------------------------ ----------
>>
>> Message: 1
>> Date: Fri, 14 Apr 2017 19:02:37 +0100
>> From: mike h <mike_k_houghton@yahoo.co.uk>
>> To: The Haskell-Beginners Mailing List - Discussion of primarily
>> beginner-level topics related to Haskell <beginners@haskell.org>
>> 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
>>
>>
>>
>>
>>
>>
>>
>> ------------------------------
>>
>> Message: 2
>> Date: Fri, 14 Apr 2017 14:17:42 -0400
>> From: David McBride <toad3k@gmail.com>
>> To: The Haskell-Beginners Mailing List - Discussion of primarily
>> beginner-level topics related to Haskell <beginners@haskell.org>
>> Subject: Re: [Haskell-beginners] Parsing
>> Message-ID:
>>
>> <CAN+Tr42ifDF62sXo6WDq32rBAPHQ+eqTkJeuk-dNr8pDfRSZXg@mail.gm >ail.com
>> 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 <mike_k_houghton@yahoo.co.uk>
>> 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
>>
>>
>> ------------------------------
>>
>> Message: 3
>> Date: Fri, 14 Apr 2017 20:35:32 +0200
>> From: Francesco Ariis <fa-ml@ariis.it>
>> 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 <$> sepEndBy1 identifier (char
>> '.')
>>
>> [1]
>> https://hackage.haskell.org/package/parsec-3.1.11/docs/Text- Parsec-Combinator.html
>>
>>
>> ------------------------------
>>
>> Message: 4
>> Date: Fri, 14 Apr 2017 20:12:14 +0100
>> From: mike h <mike_k_houghton@yahoo.co.uk>
>> To: The Haskell-Beginners Mailing List - Discussion of primarily
>> beginner-level topics related to Haskell <beginners@haskell.org>
>> Subject: Re: [Haskell-beginners] Parsing
>> Message-ID: <FF162CDE-E7E8-421B-A92E-057A643EE1A8@yahoo.co.uk >
>> Content-Type: text/plain; charset=utf-8
>>
>> Hi David,
>>
>> Thanks but I tried something like that before I posted. I’ll try again
>> maybe I mistyped.
>>
>> Mike
>> > On 14 Apr 2017, at 19:17, David McBride <toad3k@gmail.com> wrote:
>> >
>> > Try breaking it up into pieces. There a literal "package" which is
>> > dropped. There is a first identifier, then there are the rest of the
>> > identifiers (a list), then those two things are combined somehow (with
>> > :).
>> >
>> > literal "package" *> (:) <$> identifier <*> restOfIdentifiers
>> > where
>> > restOfIdentifiers :: Applicative f => f [String]
>> > restOfIdentifiers = many ((:) <$> char '.' <*> identifier
>> >
>> > I have not tested this code, but it should be close to what you are
>> > looking for.
>> >
>> > On Fri, Apr 14, 2017 at 2:02 PM, mike h <mike_k_houghton@yahoo.co.uk>
>> > wrote:
>> >> I have
>> >> data PackageDec = Pkg String deriving Show
>> >>
>> >> and a parser for it
>> >>
>> >> packageP :: Parser PackageDec
>> >> packageP = do
>> >> literal “package"
>> >> x <- identifier
>> >> xs <- many ((:) <$> 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
>> > _______________________________________________
>> > Beginners mailing list
>> > Beginners@haskell.org
>> > http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>>
>>
>>
>> ------------------------------
>>
>> Message: 5
>> Date: Fri, 14 Apr 2017 20:19:40 +0100
>> From: mike h <mike_k_houghton@yahoo.co.uk>
>> To: The Haskell-Beginners Mailing List - Discussion of primarily
>> beginner-level topics related to Haskell <beginners@haskell.org>
>> Subject: Re: [Haskell-beginners] Parsing
>> Message-ID: <D208C2B2-6E38-427D-9EAF-B9EA8532D873@yahoo.co.uk >
>> 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 <fa-ml@ariis.it> 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-Combinator.html
>> > <https://hackage.haskell.org/package/parsec-3.1.11/docs/Text >-Parsec-Combinator.html
>> > _______________________________________________
>> > 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 >
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