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Graphics.UI.Gtk.Glade
by Guenter 13 Dec '12

13 Dec '12

I am trying to compile the Gtk2Hs demo program GladeTest.hs (OS: Windows7; HaskellPlatform-2012.4.0.0-setup.exe; gtk+-bundle_2.24.10-20120208_win32; glade-3-8-3-installer.exe; I am a beginner ) The error I get is: *Prelude>*:load "GladeTest.hs" GladeTest.hs:4:8: Could not find module `Graphics.UI.Gtk.Glade' Perhaps you meant Graphics.UI.Gtk.Builder (from gtk-0.12.4) Graphics.UI.Gtk.Cairo (from gtk-0.12.4) Graphics.UI.Gtk.Gdk.GC (from gtk-0.12.4) Use -v to see a list of the files searched for. Failed, modules loaded: none. *Prelude>* I did these okay: cabal install cabal-install cabal update cabal install gtk2hs-buildtools cabal install gtk glade-3-8-3-installer.exe and I can compile and run some other Gtk2Hs demos (that do not use Glade) okay. Use -v to see a list of the files searched for. Best regards Guenter

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extended gcd
by Zbigniew Stanasiuk 11 Dec '12

11 Dec '12

Hallo everyone, I try to understand how the following recursive function works: extGCD :: Integer -> Integer -> [Integer] extGCD a 0 = [1, 0, a] extGCD a b = let (q, r) = a `quotRem` b [s, t, g] = extGCD b r in [t, s - q * t, abs g] Some hints to explain?? Regards, Z.Stanasiuk

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Beginner in Haskell
by Saadah Sabarudin 11 Dec '12

11 Dec '12

hello everyone...I'm doing my research about grid computing. I want to use haskell as part of my study to perform parallel processing. can someone tell me, what kind of tool I have to use to do parallel processing in Haskell because I'm still new in this field, so it's a little bit confusing to choose the suitable tool to use.

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Download a large file using Network.HTTP
by Cedric Fung 10 Dec '12

10 Dec '12

Hi, Are there any suggestions to download a large file with Haskell? I have read the docs for Network, Network.HTTP and Network.HTTP.Conduit, but can't find anything which fit my requirements. I want to download a large file from an HTTP URL, and show the progress instantly. Maybe some functions which read HTTP connection and return a lazy ByteString could do this work? Though I found a low-level socket lazy package, which seems to work, I just want a more high level API. Thanks and regards.

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State help request
by Joe 09 Dec '12

09 Dec '12

Hi, below is an implementation of RC4, which I started after Andrew Gwozdziewycz from NY Hack & Tell encouraged the group to write it. I mostly followed the pseudocode on Wikipedia. I'd appreciate any feedback, but I'm most distressed by the KSA generation, and would like suggestions for making either version less nasty. Thanks. import Data.Bits import Data.Vector ((!),(//),Vector,fromList) import Data.Char import Control.Monad.State type PRGA = (Vector Int,Int,Int) type Key = [Int] -- identity permutation permId :: Vector Int permId = fromList [0..255] -- generate initial PRGA ksa :: Key -> PRGA ksa key = ksaStep permId key 0 0 -- I really don't like passing the counter every time, ksaStep :: Vector Int -> Key -> Int -> Int -> PRGA ksaStep s _ 255 _ = (s,0,0) ksaStep s key i j = let j' = (j + (s!i) + (key !! (i `mod` keylength))) `mod` 256 in ksaStep (s // [(i, s!j'), (j',s!i)]) key (i+1) j' where keylength = length key -- but I tried wedging it into a State, and it's not any clearer ksa' :: Key -> PRGA ksa' key = genPRGA where genPRGA = snd $ foldl (\s a -> snd $ runState (ksaStep' a) s) (key,(permId,0,0)) [0..255] ksaStep' :: Int -> State (Key,PRGA) () ksaStep' i = do (key, (s,_,j)) <- get let j' = (j + (s!i) + (key !! (i `mod` length(key)))) `mod` 256 s' = s // [(i, s!j'), (j',s!i)] put (key,(s',i,j')) -- a round of the PRGA prgaStep :: State PRGA Int prgaStep = do (s,i,j) <- get let i' = (i + 1) `mod` 256 j' = (j + (s!i')) `mod` 256 s' = s // [(i',s!j'), (j',s!i')] put (s',i',j') return (s!((s'!i' + s'!j') `mod` 256)) keyStream :: PRGA -> [Int] keyStream p = let (i,p') = runState prgaStep p in i : keyStream p' crypt :: Key -> [Int] -> [Int] crypt k m = zipWith xor m $ keyStream $ ksa k pwCrypt :: String -> String -> [Int] pwCrypt ks ms = crypt key msg where key = map ord ks msg = map ord ms pwDecrypt :: String -> [Int] -> String pwDecrypt k c = map chr $ crypt key c where key = map ord k

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Type of function or type of the output of function?
by Trung Quang Nguyen 08 Dec '12

08 Dec '12

Hi there, I have a quickSort implementation, then I try to use :t in ghci to inspect what type I receive. I got this. *Main> :t quickSort "Hello the world" quickSort "Hello the world" :: [Char] *Main> :t quickSort [3,2, 1,-10] quickSort [3,2, 1,-10] :: (Num a, Ord a) => [a] The first one looks like :t return type of output of quickSort [Char] The second one looks like :t return the type of function (Num a, Ord a) => [a] Anybody know why this happens? When we will receive type of output, or type of function? And why (Num a, Ord a) => [a], even quickSort doesn't have explicit type, so this type comes from input type and type of function is defined in runtime? Thanks a lot! Cheers, Trung

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ANN: Haskell Beginners Google+ Community: "Haskell: The Haskell-Beginners Community"
by DekuDekuplex＠Yahoo.com 07 Dec '12

07 Dec '12

I have just created a new Haskell beginners Google+ community, "Haskell: The Haskell-Beginners Community" (see https://plus.google.com/communities/101629034466170191725) Feel free to join and contribute! -- Benjamin L. Russell -- Benjamin L. Russell / DekuDekuplex at Yahoo dot com http://dekudekuplex.wordpress.com/ Translator/Interpreter / Mobile: +011 81 90-6526-1406 "Furuike ya, kawazu tobikomu mizu no oto." -- Matsuo Basho^

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Couldn't match expected type with actual type
by Alexander _ 07 Dec '12

07 Dec '12

Hello, I have function with following signature: get :: a -> Value b I don't know concrete type of b. It can be Value Int or Value String or something else. How can i to write function only for testing? I need something like this: get :: a -> Value b get k = (Value "some_data"') but i got error: Couldn't match expected type `b' with actual type `[Char]' `b' is a rigid type variable bound by the type signature for get :: a -> Value b at Test.hs:39:8 In the first argument of `Value', namely `"some_data"' In the expression: (Value "some_data") In an equation for `get': get k = (Value "some_data") Failed, modules loaded: none. Value data declaration: data Value b = Value b deriving (Show) Thank you.

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(:
by Ezequiel Hernan Di Giorgi 06 Dec '12

06 Dec '12

this code compiles correcly: newtype State s a = State { runState :: *s* -> (a,s) } instance Monad (State s) where --return :: a -> State s a return x = State (\st -> (*x,st*)) but this one doesn't compile: newtype State s a = State { runState :: *a* -> (a,s) } instance Monad (State s) where --return :: a -> State s a return x = State (\st -> (*st,x*)) Why is this happening?

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Illegal datatype context when modelling round robin scheduler
by Xiao Jia 06 Dec '12

06 Dec '12

Hi, I'm trying to model a round robin scheduler in Haskell. > class Schedulable s where > isFinal :: s -> Bool > > class Scheduler s where > add :: (Schedulable a) => a -> s -> s > next :: (Schedulable a) => s -> (a, s) > empty :: s -> Bool > > data Schedulable a => RoundRobin = RoundRobin [a] [a] > > instance Scheduler RoundRobin where > add p (RoundRobin ps qs) = RoundRobin (ps ++ [p]) qs > > next (RoundRobin [] qs) = next (RoundRobin qs []) > next (RoundRobin (p:ps) qs) = (p, RoundRobin ps (qs ++ [p])) > > empty (RoundRobin [] _) = True > empty _ = False However, GHC complains that > main.hs:9:6: > Illegal datatype context (use -XDatatypeContexts): Schedulable a => After some searches I realized that datatype contexts are deprecated. So how can I model the scheduler without using datatype contexts? Thanks. -- Regards, Xiao Jia

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