April 2017 - Beginners

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.)

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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> >

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

18 Apr '17

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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>

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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

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will GHC optimize pattern-matching on integers?
by Patrick Pelletier 13 Apr '17

13 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

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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)

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What is the Best Haskell IDE in Windows?
by Frank Lugala 11 Apr '17

11 Apr '17

Can anyone please help?i tried to install Atom, Sublime, Emacs but i can not get them working on windows

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Same code, system, but different arch using Win32 for reading registry.
by Akos Marton 07 Apr '17

07 Apr '17

Dear Haskellers, The setup: Having a binary compiled on either x86 or x64 system (same installation, utilizing stack) in order to preserve compatibility against x64 systems. Actually, not to have 2 separate executable for each, that would be the overall goal. However it surprises me when running the x86 .exe utilizing Win32-2.5.4.1 package with ghc-8.0.2, reading out a registry key fails on x64 system with the following:/ / /me.exe: RegOpenKey: invalid argument (The system cannot find the file specified.)/ That would be fine, however the key does exists. When same code, same system, but the .exe built to be x64 it runs like a charm. A couple of question, which some of them eventually will not make sense, but still: - Can it be ghc code optimization issue? - but this is a runtime check in IO, if so, how? - Yes, I could use a built-in windows system command and parse the input of that; unless absolutely necessary I would not introduce another dependency (system package). Would love to solve it with the currently utilized weapons. - the issue just puzzles me... I would know the answer if possible. - Is it more library (Win32), ghc, binary I generate, issue? What library can I use to detect a system's architecture which works in this scenario? Another thing which convoluted in the issue... The function, /getSystemInfo :: IO SYSTEM_INFO, /can read out the underlying architecture. When compiled on x86 and run on x64 it would tell me: "I am running on x86". That's failure. Most importantly: what is the obvious I am missing? Thank you for your insights! Best, Akos Ps.: Would you/we need sample code to puzzle about I can quickly weld one. - not sure if necessary.

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State monad to help pass around game settings
by Dave Martin 07 Apr '17

07 Apr '17

I'm trying to write a game with a "settings menu" where the user can adjust gameplay options. Right now I pass all the settings around as parameters. I'm trying to figure out how to use the State monad to simplify this task, but I can't figure out how to start. Or maybe my whole design approach is wrongheaded, and not in keeping with best practices. Haskell is my first language. This is the kind of thing I have now: mainM color shape = putStrLn "\n\nMain Menu" >> (putStrLn . unlines) [ "(1) Set", "(2) Display", "(3) Quit"] >> putStr "? " >> getChar >>= \c -> case c of '1' -> set color shape '2' -> display color shape '3' -> return () _ -> mainM color shape set color shape = putStrLn "\n\nSettings" >> (putStrLn . unlines) [ "(1) Color", "(2) Shape", "(3) Main Menu"] >> putStr "? " >> getChar >>= \c -> case c of '1' -> setColor color shape '2' -> setShape color shape '3' -> mainM color shape _ -> set color shape setColor color shape = putStr ("\n\nColor is " ++ color ++ ". New color? ") >> getLine >>= \cs -> set cs shape setShape color shape = putStr ("\n\nShape is " ++ shape ++ ". New shape? ") >> getLine >>= \cs -> set color cs display color shape = putStrLn ("\n\nColor is " ++ color ++ ". Shape is " ++ shape ++ ".") >> mainM color shape

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