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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_hough...@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-0d1da19e0...@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 <toa...@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-dnr8pdfrs...@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_hough...@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...@ariis.it>
To: beginners@haskell.org
Subject: Re: [Haskell-beginners] Parsing
Message-ID: <20170414183532.ga4...@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 <$> sepEndBy1 identifier (char '.')
[1]
https://hackage.haskell.org/package/parsec-3.1.11/docs/Text-Parsec-Combinator.html
------------------------------
Message: 4
Date: Fri, 14 Apr 2017 20:12:14 +0100
From: mike h <mike_k_hough...@yahoo.co.uk>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <beginners@haskell.org>
Subject: Re: [Haskell-beginners] Parsing
Message-ID: <ff162cde-e7e8-421b-a92e-057a643ee...@yahoo.co.uk>
Content-Type: text/plain; charset=utf-8
Hi David,
Thanks but I tried something like that before I posted. I’ll try again maybe I
mistyped.
Mike
> On 14 Apr 2017, at 19:17, David McBride <toa...@gmail.com> wrote:
>
> Try breaking it up into pieces. There a literal "package" which is
> dropped. There is a first identifier, then there are the rest of the
> identifiers (a list), then those two things are combined somehow (with
> :).
>
> literal "package" *> (:) <$> identifier <*> restOfIdentifiers
> where
> restOfIdentifiers :: Applicative f => f [String]
> restOfIdentifiers = many ((:) <$> char '.' <*> identifier
>
> I have not tested this code, but it should be close to what you are looking
> for.
>
> On Fri, Apr 14, 2017 at 2:02 PM, mike h <mike_k_hough...@yahoo.co.uk> wrote:
>> I have
>> data PackageDec = Pkg String deriving Show
>>
>> and a parser for it
>>
>> packageP :: Parser PackageDec
>> packageP = do
>> literal “package"
>> x <- identifier
>> xs <- many ((:) <$> 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_hough...@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-b9ea8532d...@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...@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>
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