[Haskell-beginners] Beginners Digest, Vol 106, Issue 7
mike thomas
mjt464 at gmail.com
Wed Apr 19 04:54:17 UTC 2017
I've had good luck recently with Atom and Visual Studio code on Windows 10
and MacOS.
I'm looking at the Haskell-ghc-mod, ide-Haskell and language-Haskell
community packages right now on Atom,
and on VS Code I've got Haskell ghc-mod, Haskell syntax highlighting and
Haskell linter.
I hope this helps.
Also, do please trim stuff from previous posts not relevant to your
question. I can't do it to this reply because I'm on a phone at the
moment, and deleting that much stuff is more trouble than it's worth.
Cheers
Mike
On 19 Apr 2017 10:17 am, "Frank Lugala" <official08 at live.in> wrote:
Can anyone suggest a good Haskell IDE for windows?
------------------------------
*From:* Beginners <beginners-bounces at haskell.org> on behalf of David
McBride <toad3k at gmail.com>
*Sent:* Tuesday, April 18, 2017 4:22 PM
*To:* The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell
*Subject:* 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 <deepmindster at 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 at 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 at yahoo.co.uk>
>> To: The Haskell-Beginners Mailing List - Discussion of primarily
>> beginner-level topics related to Haskell <beginners at haskell.org>
>> Subject: [Haskell-beginners] Parsing
>> Message-ID: <2C66C9DC-30AF-41C5-B9AF-0D1DA19E0A2C at 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 at gmail.com>
>> To: The Haskell-Beginners Mailing List - Discussion of primarily
>> beginner-level topics related to Haskell <beginners at haskell.org>
>> Subject: Re: [Haskell-beginners] Parsing
>> Message-ID:
>>
>> <CAN+Tr42ifDF62sXo6WDq32rBAPHQ+eqTkJeuk-dNr8pDfRSZXg at mail.gmail.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 at 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 at haskell.org
>> > http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
Haskell-Beginners Info Page
<http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners>
mail.haskell.org
Haskell-Beginners -- The Haskell-Beginners Mailing List - Discussion of
primarily beginner-level topics related to Haskell About Haskell-Beginners
>>
>>
>> ------------------------------
>>
>> Message: 3
>> Date: Fri, 14 Apr 2017 20:35:32 +0200
>> From: Francesco Ariis <fa-ml at ariis.it>
>> To: beginners at haskell.org
>> Subject: Re: [Haskell-beginners] Parsing
>> Message-ID: <20170414183532.GA4376 at 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 <$> sepEndBy1 identifier (char
>> '.')
>>
>> [1]
>> https://hackage.haskell.org/package/parsec-3.1.11/docs/Text-
Parsec-Combinator.html
>>
>>
>> ------------------------------
>>
>> Message: 4
>> Date: Fri, 14 Apr 2017 20:12:14 +0100
>> From: mike h <mike_k_houghton at yahoo.co.uk>
>> To: The Haskell-Beginners Mailing List - Discussion of primarily
>> beginner-level topics related to Haskell <beginners at haskell.org>
>> Subject: Re: [Haskell-beginners] Parsing
>> Message-ID: <FF162CDE-E7E8-421B-A92E-057A643EE1A8 at yahoo.co.uk>
>> Content-Type: text/plain; charset=utf-8
>>
>> Hi David,
>>
>> Thanks but I tried something like that before I posted. I’ll try again
>> maybe I mistyped.
>>
>> Mike
>> > On 14 Apr 2017, at 19:17, David McBride <toad3k at gmail.com> wrote:
>> >
>> > Try breaking it up into pieces. There a literal "package" which is
>> > dropped. There is a first identifier, then there are the rest of the
>> > identifiers (a list), then those two things are combined somehow (with
>> > :).
>> >
>> > literal "package" *> (:) <$> identifier <*> restOfIdentifiers
>> > where
>> > restOfIdentifiers :: Applicative f => f [String]
>> > restOfIdentifiers = many ((:) <$> char '.' <*> identifier
>> >
>> > I have not tested this code, but it should be close to what you are
>> > looking for.
>> >
>> > On Fri, Apr 14, 2017 at 2:02 PM, mike h <mike_k_houghton at yahoo.co.uk>
>> > wrote:
>> >> I have
>> >> data PackageDec = Pkg String deriving Show
>> >>
>> >> and a parser for it
>> >>
>> >> packageP :: Parser PackageDec
>> >> packageP = do
>> >> literal “package"
>> >> x <- identifier
>> >> xs <- many ((:) <$> 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 at haskell.org
>> >> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>> > _______________________________________________
>> > Beginners mailing list
>> > Beginners at 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 at yahoo.co.uk>
>> To: The Haskell-Beginners Mailing List - Discussion of primarily
>> beginner-level topics related to Haskell <beginners at haskell.org>
>> Subject: Re: [Haskell-beginners] Parsing
>> Message-ID: <D208C2B2-6E38-427D-9EAF-B9EA8532D873 at 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 at 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 at haskell.org <mailto:Beginners at haskell.org
<Beginners at haskell.org>>
>> > http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>> > <http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners>
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>>
>> End of Beginners Digest, Vol 106, Issue 7
>> *****************************************
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