November 2014 - Beginners

abstracting parametrized record types
by Alia 29 Apr '15

29 Apr '15

Hi Folks, Here's a problem that's frustrating me: I have defined a parametrized record type which looks like this: data Job a = Job { jobID :: String , jobName :: String , jobTable :: String , jobSource :: IO [a] , jobProcessors :: [a -> a] } where a is a data model type that maps on to a database table, for example: data Person = Person { name :: String , age :: Int } deriving (Show) data Car = Car { brand :: String , value :: Double , year :: Int } deriving (Show) I would like to define a higher-level record type which contains, for configuration purposes, a certain set of jobs for execution. Let's say we call it a JobSet and which could possibly look like this: data JobSet = JobSet { jobsetID :: String , jobsetName :: String , jobs :: [Job] <-- yes I know this is not legal } Is there a legal haskell way to achieve the above objective without having to do something like this which hardcodes the model type into the jobset schema? data JobSet = JobSet { jobsetID :: String , jobsetName :: String , personJobs :: [Job Person] , carJobs :: [Job Car] } Many thanks for any enlightenment on this front. Alia

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building ghc for cross-platform raspberry pi (ubuntu)
by Miro Karpis 25 Nov '14

25 Nov '14

Hi, I'm trying to build ghc for raspberry pi, but I'm having problem with .configure. I'm following these instructions from ( https://ghc.haskell.org/trac/ghc/wiki/Building/Preparation/RaspberryPi) I have the tools installed and they should be also in my PATH,..... Please any ideas? ghc> ./configure --target=arm-linux-gnueabihf --enable-unregisterised checking for gfind... no checking for find... /usr/bin/find checking for sort... /usr/bin/sort checking for GHC version date... inferred 7.9.20141121 checking for ghc... /usr/bin/ghc checking version of ghc... 7.6.3 checking build system type... x86_64-unknown-linux-gnu checking host system type... x86_64-unknown-linux-gnu checking target system type... arm-unknown-linux-gnueabihf Build platform inferred as: x86_64-unknown-linux Host platform inferred as: x86_64-unknown-linux GHC build : x86_64-unknown-linux GHC host : x86_64-unknown-linux GHC target : arm-unknown-linux configure: Building in-tree ghc-pwd checking for path to top of build tree... /home/parallels/code/ghc checking for gcc... gcc checking whether the C compiler works... yes checking for C compiler default output file name... a.out checking for suffix of executables... checking whether we are cross compiling... no checking for suffix of object files... o checking whether we are using the GNU C compiler... yes checking whether gcc accepts -g... yes checking for gcc option to accept ISO C89... none needed checking how to run the C preprocessor... gcc -E checking for grep that handles long lines and -e... /bin/grep checking for egrep... /bin/grep -E checking for ANSI C header files... yes checking for sys/types.h... yes checking for sys/stat.h... yes checking for stdlib.h... yes checking for string.h... yes checking for memory.h... yes checking for strings.h... yes checking for inttypes.h... yes checking for stdint.h... yes checking for unistd.h... yes checking for arm-linux-gnueabihf-gcc... /home/parallels/code/tools/arm-bcm2708/gcc-linaro-arm-linux-gnueabihf-raspbian/bin/arm-linux-gnueabihf-gcc checking for arm-linux-gnueabihf-ld... /home/parallels/code/tools/arm-bcm2708/gcc-linaro-arm-linux-gnueabihf-raspbian/bin/arm-linux-gnueabihf-ld checking for arm-linux-gnueabihf-nm... /home/parallels/code/tools/arm-bcm2708/gcc-linaro-arm-linux-gnueabihf-raspbian/bin/arm-linux-gnueabihf-nm checking for arm-linux-gnueabihf-ar... /home/parallels/code/tools/arm-bcm2708/gcc-linaro-arm-linux-gnueabihf-raspbian/bin/arm-linux-gnueabihf-ar checking for arm-linux-gnueabihf-ranlib... /home/parallels/code/tools/arm-bcm2708/gcc-linaro-arm-linux-gnueabihf-raspbian/bin/arm-linux-gnueabihf-ranlib checking for llc... no checking for opt... no checking whether #! works in shell scripts... yes checking for perl... /usr/bin/perl checking version of gcc... configure: error: Need at least gcc version 3.0 (3.4+ recommended) Regards, Miro

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IntelliJ IDEA Haskell
by Andrew Bernard 20 Nov '14

20 Nov '14

Greetings List, I’d like to try IntelliJ IDEA for Haskell. There seem to be many plugins in various stages of maturity. Which one is recommended? Is this a good environment for Haskell development? Andrew

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[Haskell] ANNOUNCE: Haskell Communities and Activities Report (27th ed., November 2014)
by Mihai Maruseac 17 Nov '14

17 Nov '14

On behalf of all the contributors, we are pleased to announce that the Haskell Communities and Activities Report (27th edition, November 2014) is now available, in PDF and HTML formats: http://haskell.org/communities/11-2014/report.pdf http://haskell.org/communities/11-2014/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 April 2015: target deadline for contributions to the May 2015 edition of the HC&A 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 October (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 and Alejandro Serrano Mena <hcar at haskell.org> -- Mihai Maruseac (MM) "If you don't know, the thing to do is not to get scared, but to learn." -- Atlas Shrugged.

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Unable to sequence using 'do' operator
by Rohit Sharma 17 Nov '14

17 Nov '14

type Parser a = String -> [(a, String)] Hi, I am learning haskell and wondering why my definition of firstAndThird does not work with the do operator, however when i try to use the same using bind (firstandThird2) it works as expected. I basically want to return a tuple of first and third character of a string. Can someone please correct me what i might be overseeing? I have pasted the same code on codetidy in case if the email lose formatting. http://codetidy.com/5726/ Many Thanks, Rohit zero :: Parser a zero = \inp -> [] result :: a -> Parser a result x = \inp -> [(x, inp)] item :: Parser Char item = \inp -> case inp of [] -> [] (x:xs) -> [(x,xs)] bind :: Parser a -> (a -> Parser b) -> Parser b p `bind` f = \inp -> concat [ ((f x) inp') | (x, inp') <- p inp] sat :: (Char -> Bool) -> Parser Char sat predicate = item `bind` (\x -> if predicate x then result x else zero ) lower :: Parser Char lower = sat (\x -> 'a' <= x && x <= 'z') firstAndThird :: Parser (Char, Char) firstAndThird = do x <- item item y <- item return (x,y) firstAndThird2 :: Parser (Char, Char) firstAndThird2 = item `bind` \x -> item `bind` \y -> item `bind` \z -> result (x,z)

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Bind parser returning wrong return type
by Rohit Sharma 17 Nov '14

17 Nov '14

Hi All, Sorry if this is a very lame question but i am a beginner and much appreciate if some one could correct me? I am reading haskell book and curious why the return type of the bind operator look odd to me For the given definitions * type Parser a = String -> [(a, String)]* * item :: Parser Char* * item = \inp -> case inp of * * [] -> []* * (x:xs) -> [(x,xs)]* * bind :: Parser a -> (a -> Parser b) -> Parser b* * p `bind` f = \inp -> concat [ f x inp' | (x, inp') <- p inp]* when I define z in GHCI as * let z = item `bind` (\x -> (\y -> result (x,y))) "Rohit"* the return type is * >> :t z* * z :: Parser ([Char], Char)* Question: (1) Shouldn't the return type of (Char, [Char])? looking at the list comprehension, "(x, inp') <- p inp" should yield -> "('r', "ohit")". Next f x inp' is left associative, so f x should yield character 'r' and pass to the lambda that should return result tuple ('r', "ohit"), but why is it that z type is ([Char], char) :: (x,y) (2) How can i print the value of z in the above case on the ghci Many thanks, Rohit

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latest Haskell Platform build fails
by Michael L Martin 12 Nov '14

12 Nov '14

Hello all, I am trying to install the latest version of Haskell Platform and am having a problem. I created a new VM and installed Ubuntu 14.04 server and applied the latest updates. Also installed packages "build-essential" and "zlib1g-dev". I then downloaded and installed ghc-7.8.3 from https://www.haskell.org/ghc/download_ghc_7_8_3#x86_64linux and then downloaded and installed cabal 1.20.0.3 from http://hackage.haskell.org/package/cabal-install. After installing cabal, I did a "cabal update" followed by "cabal install hscolour". Then downloaded the latest Haskell Platform tarball (haskell-platform-2014.2.0.0-srcdist.tar.gz) and attempted to install like this: ./platform.sh ../ghc-7.8.3-x86_64-unknown-linux-deb7.tar.bz2 This fails like this: *** *** Building hptool *** Building hptool-0.1... Preprocessing executable 'hptool' for hptool-0.1... *** *** Running hptool *** # rm (for build/.markers/rel/build/target/usr/local/haskell/ghc-7.8.3-x86_64/lib/happy-1.19.4) # rm (for build/.markers/rel/build/package/happy-1.19.4/build) happy-1.19.4 needs mtl-2.1.3.1 >>> Building happy-1.19.4 # cp (for build/.markers/rel/build/package/happy-1.19.4/build) # rm (for build/.markers/rel/build/package/happy-1.19.4/build) # ghc-pkg (for build/.markers/rel/build/package/happy-1.19.4/build) # ghc-pkg (for build/.markers/rel/build/package/happy-1.19.4/build) Reading package info from "build/package/mtl-2.1.3.1/inplace.conf" ... done. mtl-2.1.3.1: package(s) with this id already exist: mtl-2.1.3.1 Error when running Shake build system: * build-all * build-product * build/product/haskell-platform-2014.2.0.0-unknown-posix-x86_64.tar.gz * build/target * build/.markers/rel/build/target/usr/local/haskell/ghc-7.8.3-x86_64/lib/happy-1.19.4 * build/.markers/rel/build/package/happy-1.19.4/build Development.Shake.command, system command failed Command: /home/mmartin/Downloads/haskell-platform-2014.2.0.0/build/ghc-bindist/local/bin/ghc-pkg register --package-db=build/package/happy-1.19.4/package.conf.d --verbose=1 build/package/mtl-2.1.3.1/inplace.conf Exit code: 1 Stderr: mtl-2.1.3.1: package(s) with this id already exist: mtl-2.1.3.1 Any suggestions on how to get past this? Thanks, Michael Martin

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Finding Keith Numbers with Haskell
by Jack Mott 12 Nov '14

12 Nov '14

I'm on day 2 of learning Haskell, and decided to write a program to find Keith Numbers as my first project. So far so good, I have some code that does the job (see below). First I would just like to ask others to look it over and let me know if I am doing anything particularly wrong or not advised here. Also, I am using the "isKeithHelper" function because I didn't know how to make use of the filter function directly with isKeith, because isKeith takes two parameters. I'm pretty sure there is a better way to handle that, any tips? Lastly, I'd like to make this as efficient as possible. At the moment I am adding [s] to the end of the list in the isKeith function, which I know is O(n) instead of O(1). I don't think there is a way to avoid either that or having to use drop which would have the same downside. Is this a case where a Sequence or Vector might be more useful? How would I convert this to use a Sequence instead? import Data.Time.Clock.POSIX import Data.Digits isKeith :: Integer -> [Integer] -> Bool isKeith n l | s == n = True | s > n = False | otherwise = isKeith n x (l ++ [s]) where s = sum l isKeithHelper :: Integer -> Bool isKeithHelper n = isKeith n (digits 10 n) main :: IO () main = do start <- getPOSIXTime print (filter isKeithHelper [1000..9999]) end <- getPOSIXTime print (end - start)

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Re: [Haskell-beginners] Are these soloutions all valid and a good use of Haskell
by Roelof Wobben 11 Nov '14

11 Nov '14

Oke, But what If I try it on a left folt. Then it must be the other way. let's say we have [1 , 2, 3] so first it's nothing and 1 is taken. so y could be one and nothuing must stay nothing because it's not the last part, The same for 2 , When it's three nothing must changed to just 3. Fold is something I find it very difficult to wrap my head around it. Maybe haskell or any functional proggramming language is not for me, Roelof Kees Bleijenberg schreef op 11-11-2014 15:46: > The program traverses the list from right to left. The first element that > acc sees during traversal is the last element of the list. > The idea is as follows: > Parameter y is the partial calcresult. > If y (in your code) is Nothing then acc returns Just x. So the new partial > result is the current list element in a Just. > The next time acc is called, y is not Nothing but a Just. In that case acc > returns the partial calcresult unchanged. > So, y is Nothing for the last list element and Just <last element of the > list> for all other list elements. > > last5 :: [a] -> Maybe a > last5 xs = foldr acc initial xs > > acc :: a -> Maybe a -> Maybe a > acc x Nothing = Just x > acc _ aJust = aJust > > initial :: Maybe a > initial = Nothing > > In your code you don’t need the ‘where’ in the acc definition. Define acc > for Nothing and for Just. > > Kees > > --------------- > Im now trying to make a rfold solution to this problem. > > Till now I have this : > > last5 :: [a] -> Maybe a > last5 xs = foldr acc initial xs > > acc :: a -> Maybe a -> Maybe a > acc x y = where > acc x Nothing = Just x > acc _ j = j > > initial :: Maybe a > initial = Nothing > > So still not working. > > Can someone explain to me in small steps how to figure out what the right > answer is to the acc part. > > Roelof > > > > > > > > > > Roelof Wobben schreef op 11-11-2014 8:20: > Thanks, this is working well. > > But just for understanding > > in the part acc x Nothing x is the input string, > and in the part _j means the same as (xs:x) > > Roelof > > > > Alex Hammel schreef op 10-11-2014 22:47: > It's an indentation issue. Make sure that there are no spaces before the bit > about 'last5 = foldr acc' etc. > > On Mon, Nov 10, 2014 at 12:48 PM, Roelof Wobben <r.wobben(a)home.nl> wrote: > Roelof Wobben schreef op 10-11-2014 20:59: > last5 :: [a] -> Maybe a > last5 = foldr acc Nothing where > acc x Nothing = Just x > acc _ j = j > > When I enter this in GHCI I see this error : > > Illegal type signature: ‘[a] -> Maybe a last5’ > > > > > _______________________________________________ > Beginners mailing list > Beginners(a)haskell.org > http://www.haskell.org/mailman/listinfo/beginners > > > > > _______________________________________________ > Beginners mailing list > Beginners(a)haskell.org > http://www.haskell.org/mailman/listinfo/beginners > > > > _______________________________________________ > Beginners mailing list > Beginners(a)haskell.org > http://www.haskell.org/mailman/listinfo/beginners >

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Re: [Haskell-beginners] Are these soloutions all valid and a good use of Haskell
by Stefan Höck 11 Nov '14

11 Nov '14

> I try to say that if the input list has only 1 item the outcome is the head > of that list. This is not how one typically thinks about folds. Look at the type of acc: acc :: a -> Maybe a -> Maybe a acc is a function wich takes two arguments (ignoring currying): On is of type `a`. This is the element type of the list you fold over. The other is of type `Maybe a`. This is the value you accumulate in your fold. The initial value is `Nothing`. You start your fold literally with nothing, since you haven't visited any element in the list yet. Now the first element of the list (see below) is passed to acc. What do you do with it? acc a Nothing = ??? a is an element in your lit, Nothing is the value you have accumulated so far. What do you do? Ignore a? Keep it? If you ignore a, you return `Nothing`. acc a Nothing = Nothing With this accumulator, the list will be traversed (ignoring lazy evaluation for the time being), all values will be ignored and your result will be Nothing, no matter what kind of list you fold over. Therefore, let's keep value `a`. In order to do that, we have to wrap it in a Just, otherwise the types won't match: acc a Nothing = Just a This function will give the right result for empty lists (Nothing) and for lists with a single value where the value is wrapped in a Just. The function will throw an error however, if you pass it a list containing more than one element, because the pattern match is not exhaustive. We need to decide what happens, when we already have a value wrapped in a Just: acc a Just b = ??? Here you have three possibilities. a) return Just b, b) return Just a, c) return Nothing. These are all the options you have. Anything else (except for throwing errors) will not compile. You can try each of the three. One will give you the head of the list, one will give you the last item of the list, and one will give you Nothing for lists with more than one element. Note that the result of your function depends on which fold you use. foldl folds from the left. The first element passed to the accumulating function will be the head of the list. foldr folds from the right. The first value passed to the accumulating function is the last in the list. So, think about the fold like this: Every element of the list is passed to your accumulator and each time you must decide whether you want to keep it or let it pass. If you keep them all, you will in the end hold the last element passed to your accumulator. If you let them all pass (except for the first, which you keep since you still have Nothing), you will end up with the first element passed to you. If you fold from the right, the first element you get is the last in the list. Keep it and don't let it go. If you fold from the left, the first element will be the head of the list. Throw them all away except for the very last one. Feel free to ask, if this doesn't make sense. Stefan > Stefan Höck schreef op 10-11-2014 15:23: > >Just a quick note: That's 'typed holes' not 'type holes' ... > > > >On Mon, Nov 10, 2014 at 03:12:20PM +0100, Stefan Höck wrote: > >>Hi Roelof > >> > >>It seems like you try to do too many things at once. Here's how you > >>could go about this step-by-step and let GHC help you implement your > >>functions along the way: > >> > >>First, give the type signature of your function: > >> > >> last5 :: [a] -> Maybe a > >> last5 = undefined > >> > >>Now, load this into GHCi or compile with GHC. If it compiles, you're > >>on the right track. Now, you want to implement it using a fold > >>(try both, foldl and foldr): > >> > >> last5 :: [a] -> Maybe a > >> last5 xs = foldr _ _ xs > >> > >>The underscores are 'type holes'. This tells the compiler to give you > >>some information about what is supposed to be placed at the two > >>positions. For the moment, we are only interested in the types of the > >>things that go there. The compiler will tell you, that > >>the hole at the first position is of type > >> > >> a -> Maybe a -> Maybe a > >> > >>and the hole at the second position is of type > >> > >> Maybe a > >> > >>Now, instead of filling the holes in place, let's define two helper > >>functions together with their type signatures. You can later on inline > >>them in your definition of last5, but for the time being, let's get as > >>much help from the compiler as we can. > >> > >> last5 :: [a] -> Maybe a > >> last5 xs = foldr acc initial xs > >> > >> acc :: a -> Maybe a -> Maybe a > >> acc = undefined > >> > >> initial :: Maybe a > >> initial = undefined > >> > >>Again, compile or load into GHCi. If you did anything wrong, the > >>compiler will tell you so. There is only one possible way to > >>implement function `initial` without cheating (= raising an error) > >> > >> initial :: Maybe a > >> initial = Nothing > >> > >>Function `acc` can be implemented in several ways. Only one of them > >>will lead to the desired behavior. Finding out the proper implementation > >>is the main point of this folding-exercise. Try also an implementation > >>using foldl. Does it behave as expected? What are the differences > >>compared to foldr? When you feed your implementations a huge list - > >>say [1..20000000] - what happens? > >> > >>Note that whenever you get an error message in a rather complex > >>function implementation, move local function definitions and lambdas > >>to the top level, give them a type signature and implement them > >>separately one at a time. Use type holes to let the compiler give > >>assistance with type signatures and possible implementations. > >>Once everything compiles and runs as expected, move the toplevel > >>definitions back to where you'd like them best. > >> > >>Stefan > >> > >>PS: A more succint implementation of last5 would use currying: > >> > >> last5 = foldr acc initial > >> > >>PPS: If you get a stack overflow with very large lists, try > >> using foldl' from Data.List (or better, once you learned > >> about the Foldable type class, from Data.Foldable). > >> > >>On Mon, Nov 10, 2014 at 02:28:08PM +0100, Roelof Wobben wrote: > >>>I tried and so far I have this : > >>> > >>>last5 = foldl(\acc x -> if x=[] then x else acc xs) [] xs > >>> > >>>But now I get parse error on the -> part. > >>> > >>>Roelof > >>> > >>> > >>>Roelof Wobben schreef op 10-11-2014 13:47: > >>>>Thanks all, > >>>> > >>>>I will try to make a fold solution as soon as I see that functional > >>>>explained in the learnyouahaskell. > >>>> > >>>>Roelof > >>>> > >>>> > >>>>Frerich Raabe schreef op 10-11-2014 11:43: > >>>>>On 2014-11-10 11:16, Karl Voelker wrote: > >>>>>>On Mon, Nov 10, 2014, at 01:50 AM, Roelof Wobben wrote: > >>>>>>>What do you experts think of the different ways ? > >>>>>>2 and 4 are quite similar, and both fine. 3 is not so good: guards > >>>>>>provide weaker guarantees than patterns, and head and tail are partial > >>>>>>functions. > >>>>>In addition to what Karl wrote, I'd like to suggest not using the > >>>>>semicolon all the time -- it's not needed and just adds noise. > >>>>> > >>>>>>All three implementations have in common that they do their own > >>>>>>recursion. It would be a good exercise to try implementing last as a > >>>>>>fold - in other words, letting the standard library do the recursion > >>>>>>for > >>>>>>you. > >>>>>Right - I suggest trying to express the problem with the most abstract > >>>>>function first, then consider using a fold, then use manual recursion. > >>>>>in > >>>>>your particular case you could exploit that for non-empty lists, getting > >>>>>the last element of a list is the same as getting the first element of > >>>>>a reversed list (and there are ready-made functions for reversing a list > >>>>>and getting the first element of a list). > >>>>> > >>>>_______________________________________________ > >>>>Beginners mailing list > >>>>Beginners(a)haskell.org > >>>>http://www.haskell.org/mailman/listinfo/beginners > >>>> > >>>_______________________________________________ > >>>Beginners mailing list > >>>Beginners(a)haskell.org > >>>http://www.haskell.org/mailman/listinfo/beginners > >>> > >_______________________________________________ > >Beginners mailing list > >Beginners(a)haskell.org > >http://www.haskell.org/mailman/listinfo/beginners > > >

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