- Beginners - Haskell.org

could not install "process-leksah" due to dependencies errors....
by André Verwijs 22 Nov '15

22 Nov '15

remote-repos: hackage.haskell.org http://hackage.haskell.org/packages/archive >> cabal install leksah process-leksah Resolving dependencies... cabal: Could not resolve dependencies: trying: leksah-0.15.1.4 (user goal) trying: hlint-1.9.23 (dependency of leksah-0.15.1.4) next goal: haskell-src-exts (dependency of leksah-0.15.1.4) rejecting: haskell-src-exts-1.16.0.1/installed-911... (conflict: hlint => haskell-src-exts>=1.17 && <1.18) rejecting: haskell-src-exts-1.17.0 (conflict: leksah => haskell-src-exts>=1.13.5 && <1.17) rejecting: haskell-src-exts-1.16.0.1, 1.16.0, 1.15.0.1, 1.15.0, 1.14.0.1, 1.14.0, 1.13.5 (conflict: hlint => haskell-src-exts>=1.17 && <1.18) rejecting: haskell-src-exts-1.13.4, 1.13.3, 1.13.2, 1.13.1, 1.13.0, 1.12.0, 1.11.1, 1.10.2, 1.10.1, 1.10.0, 1.9.6, 1.9.5, 1.9.4, 1.9.3, 1.9.0, 1.8.2, 1.8.1, 1.8.0, 1.7.2, 1.7.1, 1.7.0, 1.6.1, 1.6.0, 1.5.3, 1.5.2, 1.5.1, 1.5.0, 1.4.0, 1.3.5, 1.3.4, 1.3.3, 1.3.2, 1.3.1, 1.3.0, 1.2.0, 1.1.4, 1.1.3.1, 1.1.3, 1.1.1, 1.1.0, 1.0.1, 1.0.0, 0.5.7, 0.5.6, 0.5.5, 0.5.4, 0.5.3, 0.5.2, 0.4.8, 0.4.6, 0.4.5, 0.4.4.1, 0.4.4, 0.4.3.1, 0.4.3, 0.4.2, 0.4.1, 0.3.12, 0.3.11, 0.3.10, 0.3.9, 0.3.7, 0.3.5, 0.3.4, 0.3.3, 0.3.2, 0.2.1 (conflict: leksah => haskell-src-exts>=1.13.5 && <1.17) Backjump limit reached (change with --max-backjumps).

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NNOUNCE: Haskell Communities and Activities Report (29th ed., November 2015)
by Mihai Maruseac 22 Nov '15

22 Nov '15

On behalf of all the contributors, we are pleased to announce that the Haskell Communities and Activities Report (29th edition, November 2015) is now available, in PDF and HTML formats: http://haskell.org/communities/11-2015/report.pdf http://haskell.org/communities/11-2015/html/report.html Many thanks go to all the people that contributed to this report, both directly, by sending in descriptions, and indirectly, by doing all the interesting things that are reported. We hope you will find it as interesting a read as we did. If you have not encountered the Haskell Communities and Activities Reports before, you may like to know that the first of these reports was published in November 2001. Their goal is to improve the communication between the increasingly diverse groups, projects, and individuals working on, with, or inspired by Haskell. The idea behind these reports is simple: Every six months, a call goes out to all of you enjoying Haskell to contribute brief summaries of your own area of work. Many of you respond (eagerly, unprompted, and sometimes in time for the actual deadline) to the call. The editors collect all the contributions into a single report and feed that back to the community. When we try for the next update, six months from now, you might want to report on your own work, project, research area or group as well. So, please put the following into your diaries now: ======================================== End of March 2015: target deadline for contributions to the May 2015 edition of the HCAR Report ======================================== Unfortunately, many Haskellers working on interesting projects are so busy with their work that they seem to have lost the time to follow the Haskell related mailing lists and newsgroups, and have trouble even finding time to report on their work. If you are a member, user or friend of a project so burdened, please find someone willing to make time to report and ask them to "register" with the editors for a simple e-mail reminder in March (you could point us to them as well, and we can then politely ask if they want to contribute, but it might work better if you do the initial asking). Of course, they will still have to find the ten to fifteen minutes to draw up their report, but maybe we can increase our coverage of all that is going on in the community. Feel free to circulate this announcement further in order to reach people who might otherwise not see it. Enjoy! Mihai Maruseac <hcar at haskell.org> -- Mihai Maruseac (MM) "If you can't solve a problem, then there's an easier problem you can solve: find it." -- George Polya

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RDF with Haskell
by Jeon-Young Kang 21 Nov '15

21 Nov '15

Hello. I am a newbie learning Haskell. I'd like to import RDF in Haskell as well as to do SPARQL. I went through RDF4H, Swish, and so on. But it is still difficult to understand them for me. Anyone can suggest me to learn that? or recommend any kinds of books or tutorials? Sincerely, Jeon-Young --

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Monad for Pair
by Mike Houghton 20 Nov '15

20 Nov '15

Hi, I have newtype Pair a = P (a, a) deriving (Show) and have made Monoid, Applicative and Functor instances for it. I’m a bit stumped with Monad! instance Monad Pair where return x = P (x, x) and I got brain ache with >>= And really return x = P (x, x) doesn’t seem correct anyway. Would someone please write the Monad with an explanation? Many Thanks Mike

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Bug in solution for problem 19
by Nicolás Hertzulis 19 Nov '15

19 Nov '15

The penultimate solution for problem 19 of Ninety-Nine Haskell Problems <https://wiki.haskell.org/H-99:_Ninety-Nine_Haskell_Problems> is giving the result for the opposite of the given N. e.g. rotate "abcdefgh" (-3) gives the result for rotate "abcdefgh" 3 Here's the problem https://wiki.haskell.org/99_questions/11_to_20 and the solution https://wiki.haskell.org/99_questions/Solutions/19 I think it could be solved like this: rotate xs n | n <= 0 = (reverse . take (negate n) . reverse $ xs) ++ (reverse . drop (negate n) . reverse $ xs) | n > 0 = (drop n xs) ++ (take n xs) I know it's not great, just keeping the spirit of the original solution Is this the right list for reporting this kind of bugs? Nicolás

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Learning Haskell: Higher-order Functions
by Manuel M T Chakravarty 19 Nov '15

19 Nov '15

We have got a new tutorial chapter on ”Higher-order Functions & Combinators”: http://blog.haskellformac.com/blog/higher-order-functions-combinators Hope you enjoy it! Manuel

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Applicative Laws for the ((->) r) type
by Umair Saeed 19 Nov '15

19 Nov '15

Hi all, I'm trying to check that the Applicative laws hold for the function type ((->) r), and here's what I have so far: -- Identiy pure (id) <*> v = v pure (id) <*> v const id <*> v (\x -> const id x (g x)) (\x -> id (g x)) (\x -> g x) g x v -- Homomorphism pure f <*> pure x = pure (f x) pure f <*> pure x const f <*> const x (\y -> const f y (const x y)) (\y -> f (x)) (\_ -> f x) pure (f x) Did I perform the steps for the first two laws correctly? I'm struggling with the interchange & composition laws. For interchange, so far I have the following: -- Interchange u <*> pure y = pure ($y) <*> u u <*> pure y u <*> const y (\x -> g x (const y x)) (\x -> g x y) -- I'm not sure how to proceed beyond this point. I would appreciate any help for the steps to verify the Interchange & Composition applicative laws for the ((->) r) type. Thank you, ~Umair

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Re: [Haskell-beginners] Is there an idiom for this?
by 鲍凯文 17 Nov '15

17 Nov '15

I think i saw a relevant stackoverflow question before, but I can't find it. Anyways, the answer said something like: liftA2 (&&) pred1 pred2 which has the type a -> Bool, as desired. The reader applicative gives your input to both predicates and (&&) is applied to their results. Hope I remembered this correctly. On Nov 16, 2015 4:01 AM, <beginners-request(a)haskell.org> wrote: > 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. Is there an idiom for this? (Mark Carter) > 2. Re: Is there an idiom for this? (emacstheviking) > > > ---------------------------------------------------------------------- > > Message: 1 > Date: Mon, 16 Nov 2015 11:44:35 +0000 > From: Mark Carter <alt.mcarter(a)gmail.com> > To: beginners(a)haskell.org > Subject: [Haskell-beginners] Is there an idiom for this? > Message-ID: <5649C1A3.1010901(a)gmail.com> > Content-Type: text/plain; charset=ISO-8859-1; format=flowed > > Suppose I want to use an argument twice, as for example in the expression: > (\x -> (pred1 x) and (pred2 x)) > > Is there a shorter way of doing this? > > > ------------------------------ > > Message: 2 > Date: Mon, 16 Nov 2015 11:53:28 +0000 > From: emacstheviking <objitsu(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] Is there an idiom for this? > Message-ID: > <CAEiEuULHWhechDwUF417Rnr55RDrEVUFVOgiQ6= > 5-8jux3pXyg(a)mail.gmail.com> > Content-Type: text/plain; charset="utf-8" > > I guess it depends on the final use cases... you could use currying to > partially evaluate some stuff ready, locked and loaded as it were but the > example you have given shows to distinct functions pres1 and pred2. > > I guess the short answer is "yes" but it depends on how you do it! > > :) > Sean > > > On 16 November 2015 at 11:44, Mark Carter <alt.mcarter(a)gmail.com> wrote: > > > Suppose I want to use an argument twice, as for example in the > expression: > > (\x -> (pred1 x) and (pred2 x)) > > > > Is there a shorter way of doing this? > > _______________________________________________ > > Beginners mailing list > > Beginners(a)haskell.org > > http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners > > >

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Why this order of parameters
by martin 17 Nov '15

17 Nov '15

Hello all, there are two functions, where the order of parameters confuses me. Ons is runState :: State s a -> s -> (a, s) I understand that in the constructor s has to be first, so we can turn (State s) into a monad. But why doesn't s come first in the result too, as in runState :: State s a -> s -> (s, a) The other example is foldl, foldr foldl :: (a -> b -> a) -> a -> [b] -> a foldr :: (a -> b -> b) -> b -> [a] -> b For once the list is a list of as in foldl, but a list of as in foldr. Now that can be fixed with renaming the type parameters foldr :: (b -> a -> a) -> a -> [b] -> a this is the exact same thing and resembles more the type of foldl. But still the function argument has its parameters flipped. foldl's function takes the list element as second parameter and foldr's takes it as first parameter. Why is that so?

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Is there an idiom for this?
by Mark Carter 17 Nov '15

17 Nov '15

Suppose I want to use an argument twice, as for example in the expression: (\x -> (pred1 x) and (pred2 x)) Is there a shorter way of doing this?

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