- Beginners - Haskell.org

Re: [Haskell-beginners] Foldable for (,)
by Jonathon Delgado 23 Apr '17

23 Apr '17

So then the question becomes, why do tuples need Foldable if the functions it defines aren't useful?

1 0

Traversable Instance of the Constant type - Think I got it
by Atrudyjane 21 Apr '17

21 Apr '17

Got confused thinking that the pure function called by Constant's traversable instance was the pure defined in Constant's applicative, which just returns Constant. Was explained to me that the type is inferred by the 'f' passed to it. If you read my OP, thank you. Regards, Andrea Sent with [ProtonMail](https://protonmail.com) Secure Email.

1 0

Traversable Instance of the Constant type
by Atrudyjane 20 Apr '17

20 Apr '17

Pretty sure I'm getting bogged down on something simple, but how does the traverse function actually embed the Constant type in the f context passed to it? Or is it the pure defined in the f context that's being called? Thank you, Andrea newtype Constant a b = Constant { getConstant :: a } deriving (Eq, Show) instance Functor (Constant a) where fmap f (Constant a) = Constant a instance Monoid a => Applicative (Constant a) where pure a = Constant { getConstant = mempty } Constant a <*> Constant b = Constant { getConstant = a `mappend` b } instance Foldable (Constant a) where foldMap f (Constant a) = mempty instance Traversable (Constant a) where traverse f (Constant a) = pure $ Constant a λ> traverse (\x -> Just x) (Constant 1) Just (Constant {getConstant = 1}) Sent with [ProtonMail](https://protonmail.com) Secure Email.

1 0

Setup.hs woes
by Ramnath R Iyer 19 Apr '17

19 Apr '17

Hello, I've been having some trouble with Setup.hs and could use some help. I'm using stack for building my project, and would like to auto-generate an HTML file as part of the build process. I'm able to specify hooks in Setup.hs to do this, but I'm running into problems. 1. I'm not sure what the "correct" or suitable paths are for publishing HTML documentation. The closest I could find was the doc dir on the install path (see below). Should the HTML be created in a built folder and later copied over to an install/dist dir? The semantics associated with each of these is not clear to me. 2. I played around with some hooks. I was able to generate the HTML file in the installation's doc dir, but it looks like that dir is only created as part of the install hook (naturally), whereas the install hook is never triggered by `stack build`. (The code below never executes if I use the instHook.) What is the right hook to use, and how? import Distribution.PackageDescription (PackageDescription) import Distribution.Simple (defaultMainWithHooks, simpleUserHooks) import Distribution.Simple.InstallDirs (InstallDirs(..), docdir) import Distribution.Simple.LocalBuildInfo (LocalBuildInfo(..), installDirTemplates, fromPathTemplate) import Distribution.Simple.Setup (InstallFlags) import Distribution.Simple.UserHooks (UserHooks, instHook) import System.Exit (ExitCode(..)) import System.Process (system) main :: IO () main = defaultMainWithHooks simpleUserHooks { instHook = readme } -- does not get invoked; postBuild does work, but where to put the generated file? readme :: PackageDescription -> LocalBuildInfo -> UserHooks -> InstallFlags -> IO () readme _ (LocalBuildInfo { installDirTemplates = InstallDirs { docdir = docdir' }}) _ _ = do putStrLn "Generating README.html from README.md..." exitCode <- system $ "./doc/generate " ++ (show destination) case exitCode of ExitSuccess -> return () ExitFailure _ -> fail "README.html could not be generated from README.md" return () where destination = fromPathTemplate docdir' *** (Apologies if this is a duplicate email, wasn't sure if this already got sent previously.)

1 0

Re: [Haskell-beginners] Beginners Digest, Vol 106, Issue 7
by Andrey Klaus 19 Apr '17

19 Apr '17

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(a)haskell.org>: > Send Beginners mailing list submissions to > beginners(a)haskell.org > > To subscribe or unsubscribe via the World Wide Web, visit > http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners > or, via email, send a message with subject or body 'help' to > beginners-request(a)haskell.org > > You can reach the person managing the list at > beginners-owner(a)haskell.org > > When replying, please edit your Subject line so it is more specific > than "Re: Contents of Beginners digest..." > > > 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(a)yahoo.co.uk> > To: The Haskell-Beginners Mailing List - Discussion of primarily > beginner-level topics related to Haskell <beginners(a)haskell.org> > Subject: [Haskell-beginners] Parsing > Message-ID: <2C66C9DC-30AF-41C5-B9AF-0D1DA19E0A2C(a)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(a)gmail.com> > To: The Haskell-Beginners Mailing List - Discussion of primarily > beginner-level topics related to Haskell <beginners(a)haskell.org> > Subject: Re: [Haskell-beginners] Parsing > Message-ID: > <CAN+Tr42ifDF62sXo6WDq32rBAPHQ+eqTkJeuk-dNr8pDfRSZXg@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(a)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(a)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(a)ariis.it> > To: beginners(a)haskell.org > Subject: Re: [Haskell-beginners] Parsing > Message-ID: <20170414183532.GA4376(a)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 <$> sepEndBy1 identifier (char '.') > > [1] https://hackage.haskell.org/package/parsec-3.1.11/docs/ > Text-Parsec-Combinator.html > > > ------------------------------ > > Message: 4 > Date: Fri, 14 Apr 2017 20:12:14 +0100 > From: mike h <mike_k_houghton(a)yahoo.co.uk> > To: The Haskell-Beginners Mailing List - Discussion of primarily > beginner-level topics related to Haskell <beginners(a)haskell.org> > Subject: Re: [Haskell-beginners] Parsing > Message-ID: <FF162CDE-E7E8-421B-A92E-057A643EE1A8(a)yahoo.co.uk> > Content-Type: text/plain; charset=utf-8 > > Hi David, > > Thanks but I tried something like that before I posted. I’ll try again > maybe I mistyped. > > Mike > > On 14 Apr 2017, at 19:17, David McBride <toad3k(a)gmail.com> wrote: > > > > Try breaking it up into pieces. There a literal "package" which is > > dropped. There is a first identifier, then there are the rest of the > > identifiers (a list), then those two things are combined somehow (with > > :). > > > > literal "package" *> (:) <$> identifier <*> restOfIdentifiers > > where > > restOfIdentifiers :: Applicative f => f [String] > > restOfIdentifiers = many ((:) <$> char '.' <*> identifier > > > > I have not tested this code, but it should be close to what you are > looking for. > > > > On Fri, Apr 14, 2017 at 2:02 PM, mike h <mike_k_houghton(a)yahoo.co.uk> > wrote: > >> I have > >> data PackageDec = Pkg String deriving Show > >> > >> and a parser for it > >> > >> packageP :: Parser PackageDec > >> packageP = do > >> literal “package" > >> x <- identifier > >> xs <- many ((:) <$> 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(a)haskell.org > >> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners > > _______________________________________________ > > Beginners mailing list > > Beginners(a)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(a)yahoo.co.uk> > To: The Haskell-Beginners Mailing List - Discussion of primarily > beginner-level topics related to Haskell <beginners(a)haskell.org> > Subject: Re: [Haskell-beginners] Parsing > Message-ID: <D208C2B2-6E38-427D-9EAF-B9EA8532D873(a)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(a)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(a)haskell.org <mailto:Beginners@haskell.org> > > http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners < > http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners> >

5 4

how does hgearman-client work?
by info＠maximka.de 18 Apr '17

18 Apr '17

2 6

Conditionals where more than one case is true
by PATRICK BROWNE 17 Apr '17

17 Apr '17

I am having difficulty in evaluating conditionals where more than one case is true. I require a program to express the following conditions: 1) x is faster than y if x is a buffolo and y is a pig. 2) x is faster than y if x is a pig and y is a slug. Both of these conditions are true (expressed in fact predicate). I include my attempt using guards (I have also tried case/if-then-else) I can apreciate that the program works when the first guard condition is met (faster Bob Joe). I am not sure why it fails on the second case (faster Bob Joe) Can I program the requirement using conditionals? Can I pattern match is some way? Is there another way to encode the requirement. Regards, Pat data E = Bob | Joe | Steve | Buffalo | Pig | Slug deriving Show fact Buffalo Bob = True fact Pig Joe = True fact Slug Steve = True faster x y | fact Buffalo x && fact Pig y = True | fact Pig x && fact Slug y = True | otherwise = False -- faster Bob Joe OK -- faster Steve Bob ***Exception: Faster.hs:(5,1)-(7,23): Non-exhaustive patterns in function fact -- This email originated from DIT. If you received this email in error, please delete it from your system. Please note that if you are not the named addressee, disclosing, copying, distributing or taking any action based on the contents of this email or attachments is prohibited. www.dit.ie Is ó ITBÁC a tháinig an ríomhphost seo. Má fuair tú an ríomhphost seo trí earráid, scrios de do chóras é le do thoil. Tabhair ar aird, mura tú an seolaí ainmnithe, go bhfuil dianchosc ar aon nochtadh, aon chóipeáil, aon dáileadh nó ar aon ghníomh a dhéanfar bunaithe ar an ábhar atá sa ríomhphost nó sna hiatáin seo. www.dit.ie Tá ITBÁC ag aistriú go Gráinseach Ghormáin – DIT is on the move to Grangegorman <http://www.dit.ie/grangegorman>

2 2

Parsing
by mike h 14 Apr '17

14 Apr '17

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

3 6

will GHC optimize pattern-matching on integers?
by Patrick Pelletier 14 Apr '17

14 Apr '17

Suppose I am doing a pattern match on a large number of consecutive (or mostly-consecutive) integers: foo 0 = something foo 1 = somethingElse ... foo 1000 = anotherThing Will GHC optimize this to a table lookup, or is it going to test each integer in turn? Am I better off using a Vector or Map instead of pattern matching? Thanks, --Patrick

2 2

Re: [Haskell-beginners] What is the Best Haskell IDE in Windows?
by Gregory Guthrie 12 Apr '17

12 Apr '17

I like the HEAT simple IDE for small one-file programs, which are common in introductory classes. (We made a few local changes and additions - happy to share.) I also like using IntelliJ IDEA with Haskell plugin, and that is probably the most polished and best actual IDE. (The eclipse plugin is abandoned and depreciated.) For larger projects, Leksah is good, although the documentation is sparse (IMHO), and it has a bit of a learning/setup curve. ---------------------------------------------------------------- -----Original Message----- Sent: Wednesday, April 12, 2017 7:00 AM Subject: Beginners Digest, Vol 106, Issue 4 1. What is the Best Haskell IDE in Windows? (Frank Lugala)

1 0