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FFI foreignPtr construction
by PICCA Frederic-Emmanuel 27 Jan '16

27 Jan '16

Hello, since the last time I think that I understand how to manage Ptr, now I woud like to masterize the ForeignPtr in order to let haskell managed the life of my C objects. So I try to create a Geoemtry object like this. -- data Geometry newtype Geometry = Geometry (Ptr Geometry) deriving (Show, Storable) newGeometry :: Factory -> ForeignPtr Geometry newGeometry f = unsafePerformIO $ do geometry <- c_hkl_factory_create_new_geometry f newForeignPtr c_hkl_geometry_free geometry foreign import ccall safe "hkl.h hkl_factory_create_new_geometry" c_hkl_factory_create_new_geometry :: Factory -> IO (Geometry) foreign import ccall safe "hkl.h &hkl_geometry_free" c_hkl_geometry_free :: FunPtr (Geometry -> IO ()) the C signature are HKLAPI HklGeometry *hkl_factory_create_new_geometry(const HklFactory *self) HKL_ARG_NONNULL(1); HKLAPI void hkl_geometry_free(HklGeometry *self) HKL_ARG_NONNULL(1); But when I try to compile this code, I get this error message 1 of 1] Compiling Hkl.C ( src/Hkl/C.hs, dist/build/Hkl/C.o ) src/Hkl/C.hs:51:33: Couldn't match type ‘Geometry’ with ‘Ptr Geometry’ Expected type: GHC.ForeignPtr.FinalizerPtr Geometry Actual type: FunPtr (Geometry -> IO ()) In the first argument of ‘newForeignPtr’, namely ‘c_hkl_geometry_free’ In a stmt of a 'do' block: newForeignPtr c_hkl_geometry_free geometry src/Hkl/C.hs:51:53: Couldn't match expected type ‘Ptr Geometry’ with actual type ‘Geometry’ In the second argument of ‘newForeignPtr’, namely ‘geometry’ In a stmt of a 'do' block: newForeignPtr c_hkl_geometry_free geometry I do not understand what is wrong in my code thanks if you can help Frederic

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Parsing 'A's and then ('A's or 'B's)
by Simon Jakobi 27 Jan '16

27 Jan '16

Hi! I want to test whether a sequence of the characters 'A' and 'B' can represent a sequence of the symbols x and y where x may be represented by one or more 'A's and y may be represented by one or more 'A's or one or more 'B's. In code, I would like to see the following: λ> "AABB" `represents` [x, y] True λ> "AA" `represents` [x, y] True But with my current implementation using attoparsec only the first example works as expected: import Control.Applicative import Data.Attoparsec.ByteString.Char8 import Data.ByteString import Data.Either import Data.Foldable import Data.Word type Symbol = Parser [Word8] x :: Symbol x = many1 (char8 'A') y :: Symbol y = many1 (char8 'A') <|> many1 (char8 'B') represents :: ByteString -> [Symbol] -> Bool bs `represents` symbols = isRight $ parseOnly ((sequenceA_ symbols) *> endOfInput) bs It seems that in "AA" `represents` [x, y] x consumes all the 'A's, leaving none for y. Is it possible to solve this with attoparsec or are there other parsing libraries that I could use instead? Cheers, Simon

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Ambigous type variable, why this error?
by martin 27 Jan '16

27 Jan '16

Hello all, here is something where I don't understand the second error: *Main> (Open [1,2,3]) <: (Open ([1,2,4])) <interactive>:94:8: No instance for (Num a0) arising from the literal ‘1’ The type variable ‘a0’ is ambiguous Note: there are several potential instances: instance Num Double -- Defined in ‘GHC.Float’ instance Num Float -- Defined in ‘GHC.Float’ instance Integral a => Num (GHC.Real.Ratio a) -- Defined in ‘GHC.Real’ ...plus 46 others In the expression: 1 In the first argument of ‘Open’, namely ‘[1, 2, 3]’ In the first argument of ‘(<:)’, namely ‘(Open [1, 2, 3])’ Okay, I understand this one, but why this: <interactive>:94:16: No instance for (Poset a0) arising from a use of ‘<:’ The type variable ‘a0’ is ambiguous Note: there are several potential instances: instance (Eq a, Ord a, Poset a) => Poset (Crust a) -- <== yes, yes, yes, take this one -- Defined at /home/martin/projects/haskell/currychicken/opal/Poset.hs:83:10 instance (Eq a, Ord a, Poset a) => Poset (PsSet a) -- Defined at /home/martin/projects/haskell/currychicken/opal/Poset.hs:50:10 instance (Eq a, Ord a, Poset a) => Poset (PsList a) -- Defined at /home/martin/projects/haskell/currychicken/opal/Poset.hs:46:10 ...plus one other In the expression: (Open [1, 2, 3]) <: (Open ([1, 2, 4])) In an equation for ‘it’: it = (Open [1, 2, 3]) <: (Open ([1, 2, 4])) The operands of (<:) are clearly Crusts, so (PsSet a) or (PsList a) shouldn't be options *Main> :t Open [1,2,3] Open [1,2,3] :: Num a => Crust a *Main> The problem goes away, when I make sure my list elements are Ints *Main> (Open [1::Int,2,3]) <: (Open ([1,2,4])) False But why do I see the second error at all? Here is the complete code: {-# Language FlexibleInstances #-} {-# Language UndecidableInstances #-} import qualified Data.List as L import qualified Data.Set as S import Debug.Trace import Test.QuickCheck hiding ((==>)) ------------------------------------------------------------ class Poset p where ------------------------------------------------------------ (<:) :: p -> p -> Bool instance Poset Int where (<:) = (==) ------------------------------------------------------------ -- Alternatives ------------------------------------------------------------ newtype PsList a = PsList [a] newtype PsSet a = PsSet (S.Set a) isSubPolist :: (Poset a) => [a] -> [a] ->Bool isSubPolist as bs = all includedInBs as where includedInBs a = any (a <:) bs instance (Eq a, Ord a, Poset a) => Poset (PsList a) where (PsList as) <: (PsList bs) = isSubPolist as bs instance (Eq a, Ord a, Poset a) => Poset (PsSet a) where (PsSet as) <: (PsSet bs) = isSubPolist (S.toList as) (S.toList bs) ------------------------------------------------------------ data Crust a = Open [a] | Closed [a] ------------------------------------------------------------ deriving (Eq, Ord, Show) instance (Eq a, Ord a, Poset a) => Poset (Crust a) where (<:) (Open as) (Closed bs) = False (<:) (Closed as) (Closed bs) = as == bs (<:) (Open _) (Open []) = True (<:) (Open []) (Open _) = False (<:) (Open (x:xs)) (Open (y:ys)) = x <: y && (Open xs) <: (Open ys) (<:) (Closed _) (Open []) = True (<:) (Closed []) (Open _) = False (<:) (Closed (x:xs)) (Open (y:ys)) = x <: y && (Closed xs) <: (Open ys)

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compilation error for jhc
by Fabien R 25 Jan '16

25 Jan '16

Hello, I want to build jhc 0.82 on debian/amd64. I downloaded the sources and successfully configured the build. The compilation fails with this error: [149 of 186] Compiling Ho.Binary ( src/Ho/Binary.hs, src/Ho/Binary.o ) src/Ho/Binary.hs:164:10: Duplicate instance declarations: instance [overlap ok] Binary Data.Version.Version -- Defined at src/Ho/Binary.hs:164:10 instance [safe] Binary Data.Version.Version -- Defined in `binary-0.8.0.0:Data.Binary.Class' Makefile:1441: recipe for target 'jhc' failed The corresponding source code in src/Ho/Binary.hs is: import Data.Binary import qualified Data.Version instance Binary Data.Version.Version where put (Data.Version.Version a b) = put a >> put b get = liftM2 Data.Version.Version get get In the package binary, is this error due to this line ? import Data.Version (Version(..)) Any hint to solve this ? -- Fabien

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Custom type classes
by Daniel Hinojosa 25 Jan '16

25 Jan '16

I am pretty sure I have a good handle on type classes, but this one is perplexing me in Haskell. I created custom Tuples called Tuple2 and Tuple3 (I know Haskell already has Tuples, just thought I would create my own for this exercise). Then I wanted to create a type class that would have a method called first that would get the first element of the tuple regardless of what kind of Tuple it is. Tuple2, Tuple3, etc. Here is what I have: data Tuple3 a b c = Tuple3 a b c deriving (Show) data Tuple2 a b = Tuple2 a b deriving (Show) class Indexable idx where first :: idx c -> a instance Indexable (Tuple2 a) where first (Tuple2 a b) = a In my main, I try to get call putStrLn $ show $ first $ Tuple2 1 "One" I was greeted with the following trace: Couldn't match expected type ‘a1’ with actual type ‘a’ ‘a’ is a rigid type variable bound by the instance declaration at TypeClasses.hs:35:10 ‘a1’ is a rigid type variable bound by the type signature for first :: Tuple2 a c -> a1 at TypeClasses.hs:36:4 Relevant bindings include a :: a (bound at TypeClasses.hs:36:18) first :: Tuple2 a c -> a1 (bound at TypeClasses.hs:36:4) In the expression: a In an equation for ‘first’: first (Tuple2 a b) = a Help is appreciated.

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Functional Programming for the Object Oriented
by Øystein Kolsrud 24 Jan '16

24 Jan '16

I gave a presentation on functional programming a couple of weeks ago that I thought might be of interest to this community as well: https://www.youtube.com/watch?v=I2tMmsZC1ZU It's target audience is programmers familiar with object oriented programming, and it presents how concepts from the functional paradigm are relevant also in most modern OO languages. For more information about the presentation, please refer to the following site: http://www.foocafe.org/previous_event/functional-programming-for-the-object… -- Mvh Øystein Kolsrud

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FFI and opac struct
by PICCA Frederic-Emmanuel 24 Jan '16

24 Jan '16

Hello Here the code I try to use I get an array of (HklFactory *) via the GetAll method and I want to read for each of these HklFactory the name of the factory with the NameGet method. -- data HklFactory newtype HklFactory = HklFactory (Ptr HklFactory) deriving (Show, Storable) hklFactoryGetAll :: IO [HklFactory] hklFactoryGetAll = alloca $ \ptr -> do factories <- c_hkl_factory_get_all ptr n <- peek ptr peekArray n factories foreign import ccall unsafe "hkl.h hkl_factory_get_all" c_hkl_factory_get_all :: Ptr Int -> IO (Ptr HklFactory) hklFactoryNameGet :: Ptr HklFactory -> IO String hklFactoryNameGet factory = do name <- c_hkl_factory_name_get factory peekCString name foreign import ccall unsafe "hkl.h hkl_factory_name_get" c_hkl_factory_name_get :: Ptr HklFactory -> IO CString main :: IO () main = do initGUI factories <- hklFactoryGetAll names <- mapM hklFactoryNameGet factories print factories print names the signature of the C method are: typedef struct _HklFactory HklFactory; HKLAPI HklFactory **hkl_factory_get_all(size_t *n) HKL_ARG_NONNULL(1); HKLAPI const char *hkl_factory_name_get(const HklFactory *self) HKL_ARG_NONNULL(1); when I run the haskell code, I just get garbage when I try to extract the name of all HklFActoies here the output [HklFactory 0xb777b920,HklFactory 0xb777ba6c,HklFactory 0xb777ba98,HklFactory 0xb777bac4,HklFactory 0xb777bbf8,HklFactory 0xb777bef8,HklFactory 0xb777c4f0,HklFactory 0xb777c524,HklFactory 0xb777c6c0,HklFactory 0xb777c784,HklFactory 0xb777c8a8,HklFactory 0xb777c8dc,HklFactory 0xb777c908,HklFactory 0xb777c9cc] [",v+v<w\STX","2v,vw\EOT","2v|.vw\EOT","2v\DC40vw\EOT","5v\\3v\DC4w\EOT","_8v86v\DC4w\ACK","z@v9v\fw\ACK","@v\EOT<v@w\ACK","EvBvw\ACK","GvEvw\ENQ","yNvGvw\ACK","NvHJvw\EOT","Nv,Lv$w\ENQ","PvNvw\EOT"] Instead of human readable names I get this garbage. So I do not understand what is going on and what is wrong in my haskell code. thanks for your help Frederic

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FW: Beginners Digest, Vol 91, Issue 26
by pmcilroy＠gmail.com 22 Jan '16

22 Jan '16

Don’t worry about the time to append rather than pretend, it is not “exponential” nor even cubic. You are making a list of length in an N^2 algorithm. Appending to a list adds another tiny O(N^2) step, which is dominated by the main algorithm (also O(N^2), which both takes longer and evaluates vastly more values (failures as well as successes.) Message: 3 Date: Thu, 21 Jan 2016 19:05:17 +0000 From: Rein Henrichs <rein.henrichs(a)gmail.com> To: The Haskell-Beginners Mailing List - Discussion of primarily beginner-level topics related to Haskell <beginners(a)haskell.org>, masonmlai(a)gmail.com Subject: Re: [Haskell-beginners] program not running lazily Message-ID: <CAJp6G8zemeHpYwZ36ShMJ-m_BishVnUJpoqEmK=DosKHRVaqeg(a)mail.gmail.com> Content-Type: text/plain; charset="utf-8" s/not/note, sorry On Thu, Jan 21, 2016 at 10:42 AM Rein Henrichs <rein.henrichs(a)gmail.com> wrote: > But not that doing so will cause the program to have an exponential > runtime as each new ys must be repeatedly traversed to append a [y].. The > alternative is to *unfold* your list by recursing on the right hand side of > the (:) to add new elements. > > On Thu, Jan 21, 2016 at 5:43 AM Doug McIlroy <doug(a)cs.dartmouth.edu> > wrote: > >> Each time you find another good 9-mer, you add it to >> the head of the list. This means that the ultimate >> list will be in reverse order of discovery: the first element >> to be printed is the last one to be found. To get >> output in the order it was discovered, build the >> output by ys++[y] rather than y:ys. >> _______________________________________________ >> Beginners mailing list >> Beginners(a)haskell.org >> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners >> >

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program not running lazily
by Doug McIlroy 21 Jan '16

21 Jan '16

Each time you find another good 9-mer, you add it to the head of the list. This means that the ultimate list will be in reverse order of discovery: the first element to be printed is the last one to be found. To get output in the order it was discovered, build the output by ys++[y] rather than y:ys.

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program not running lazily
by Mason Lai 21 Jan '16

21 Jan '16

Hi, I'm teaching myself Haskell and attempting to use it for day-to-day tasks at work. I signed up for this list a few weeks ago, and this is my first post. Here's my problem: I'm given a list of 45 Strings, each nine Chars long. The only characters are 'A', 'C', 'G' and 'T'. (This is for a bioinformatics application.) I'll refer to any nine-length String composed of these Chars as a "9-mer". I need to generate a larger list of 9-mers such that no two 9-mers have a Levenshtein distance of less than 3. (The list I start with satisfies this requirement.) I can generate as many 9-mers as I possibly can, but this process is very slow. It's also not being computed lazily; calling head on the output list forces the entire list of 9-mers to be computed. *I'd like to understand why this list isn't being computed lazily, and how I can change my code to make it so. *My knowledge of monads is pretty poor as well, so a digression or a series of questions about why the line [9] >>= (`replicateM` "ACGT") works would also be helpful, as would general tips about writing clean code. This is an O(n^2) operation, so I'm not expecting it to be slow. However, I figured that I'd just be able to take the first N elements from the output list. Here's what I have: import Control.Monad main :: IO () main = interact processData processData :: String -> String processData = unlines . (`merge` ([9] >>= (`replicateM` "ACGT"))) . lines -- Merges two lists of things into a single list merge :: Eq a => [[a]] -> [[a]] -> [[a]] merge xs [] = xs merge xs ys = merge ((head ys) `addInto` xs) $ tail ys -- Adds a thing into a list if it is "different" enough addInto :: Eq a => [a] -> [[a]] -> [[a]] y `addInto` ys = if ((minimum $ map (dist y) ys) < 3) then ys else y:ys -- Calculate the Levenshtein distance -- Lloyd Allison algorithm. See Haskell wiki -- code omitted dist :: Eq a => [a] -> [a] -> Int My workaround to getting a smaller subset of the whole list is to replace [9] >>= (`replicateM` "ACGT") with take 10000 $ [9] >>= (`replicateM` "ACGT") Thanks! Mason Lai

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