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running HGL on Mac OSX
by Vojtěch Knyttl 13 Feb '10

13 Feb '10

Hello, I am having difficulties running HGL on Mac OSX, Snow Leopard. Here is what I've done: –– 1. Installed X11 for OSX ("About" dialog says: The X Window System, XQuartz 2.3.4 (xorg-server 1.4.2-apple45)) –– 2. Got X11 from Hackage – configured, built and installed – however, there where quite a lot of warnings: … [ 8 of 25] Compiling Graphics.X11.Xlib.Color ( Graphics/X11/Xlib/Color.hs, dist/build/Graphics/X11/Xlib/Color.o ) Graphics/X11/Xlib/Color.hs:38:0: Warning: The import of `Foreign.C.Types' is redundant except perhaps to import instances from `Foreign.C.Types' To import instances alone, use: import Foreign.C.Types() [ 9 of 25] Compiling Graphics.X11.Xlib.Font ( dist/build/Graphics/X11/Xlib/Font.hs, dist/build/Graphics/X11/Xlib/Font.o ) Graphics/X11/Xlib/Font.hsc:167:1: Warning: A do-notation statement discarded a result of type CInt. Suppress this warning by saying "_ <- xTextExtents font_struct c_string (fromIntegral nchars) direction_return font_ascent_return font_descent_return overall_return", or by using the flag -fno-warn-unused-do-bind … and so on. –– 3. Got HGL package, built and installed with some warnings like: … [ 1 of 27] Compiling Graphics.HGL.Internals.Flag ( Graphics/HGL/Internals/Flag.hs, dist/build/Graphics/HGL/Internals/Flag.o ) Graphics/HGL/Internals/Flag.hs:1:0: Warning: Module `Prelude' is deprecated: You are using the old package `base' version 3.x. Future GHC versions will not support base version 3.x. You should update your code to use the new base version 4.x. … –– Now, when I try to run simple program like: main = runGraphics $ withWindow_ "Hello World Window" (300, 200) $ \ w -> do drawInWindow w $ text (100, 100) "Hello World" drawInWindow w $ ellipse (100, 80) (200, 180) getKey w it says: $ ghci graphics.hs GHCi, version 6.12.1: http://www.haskell.org/ghc/ :? for help Loading package ghc-prim ... linking ... done. Loading package integer-gmp ... linking ... done. Loading package base ... linking ... done. Loading package ffi-1.0 ... linking ... done. Ok, modules loaded: Main. Prelude Main> main Loading package syb-0.1.0.2 ... linking ... done. Loading package X11-1.5.0.0 ... can't load .so/.DLL for: X11 (dlopen(libX11.dylib, 9): image not found) Prelude Main> Just to precise, I have X11 running (it can be seen in the OSX dock) Thanks

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Regular expressions in case
by legajid 12 Feb '10

12 Feb '10

Hi, is it possible to write regular expressions in case .. of ? I tried it, with no success. a x = case x of "a1" -> "A1" "a2" -> "A2" "a*" -> "A*" "b1" -> "B1" _ -> "_" for x="a3", result is _, instead of A* that i wanted. Thks, Didier

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Data.Tree computation
by Gabi 12 Feb '10

12 Feb '10

Hi guys, I've tried to use Data.Tree as a computation tree (each node is numerical function, each leaf is a terminal) It kinda works, but the code seems very verbose. How can it made more concise ? I am sure I missed a lot of shortcuts and idioms. -- file t.hs import qualified Data.Tree as T data Term = TInt Int| TDouble Double deriving (Show, Eq) data Func = Plus | Minus | Mult | Div deriving (Show, Eq) data ANode = GFunc Func | GTerm Term deriving (Show, Eq) fNode :: Func -> T.Forest ANode-> T.Tree ANode fNode f = T.Node (GFunc f) tNode:: Term -> T.Tree ANode tNode t = T.Node (GTerm t) [] calc :: T.Tree ANode -> Double calc (T.Node (GTerm (TInt n))[]) = fromIntegral n :: Double calc (T.Node (GFunc Plus) xs ) = foldl1 (+) (map calc xs) calc (T.Node (GFunc Minus) xs ) = foldl1 (-) (map calc xs) calc (T.Node (GFunc Mult) xs ) = foldl1 (*) (map calc xs) calc (T.Node (GFunc Div) xs ) = foldl1 (/) (map calc xs) -- (/ (+ 5 5 (- 10 100)) 10) - calc Should return -8.0 aTree = fNode Div [fNode Plus [tNode $ TInt 5,tNode $ TInt 5, fNode Minus [tNode $ TInt 10,tNode $ TInt 100]], tNode (TInt 10)] Regards, Gabi http://bugspy.net

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ghc problem
by jean verdier 11 Feb '10

11 Feb '10

Trying to compile sources gives me: ghc: internal error: evacuate(static): strange closure type 34 (GHC version 6.10.4 for i386_unknown_linux) Please report this as a GHC bug: http://www.haskell.org/ghc/reportabug This already occured to me some times before and a cleanup and recompile would work. This time i am stuck as i can't compile anymore. The ghc version i use is the 6.10.4 from the ghc home page binary distribution i think. uname -a gives : Linux localhost.localdomain 2.6.29.4-167.fc11.i586 #1 SMP Wed May 27 17:14:37 EDT 2009 i686 athlon i386 GNU/Linux The command line is: ghc -fwarn-incomplete-patterns -fno-warn-unused-binds -fno-warn-unused-matches -fno-warn-unused-imports -fno-warn-missing-signatures -fno-glasgow-exts -Werror -outputdir /tmp/mfk/build -package ghc -i/tmp/mfk/gensrc --make common/MFK/Common/SrcGen.hs common/MFK/Common/Conf.hs Any hints ?

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HackageDB - How to browse by development status
by Gabi 11 Feb '10

11 Feb '10

Thanks Stephen This is really useful link. At least now I can guess what package is "popular" what is esoteric. -- Regards, Gabi http://bugspy.net

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HackageDB - How to browse by development status
by Gabi 10 Feb '10

10 Feb '10

Hi I really miss the option browsing this big DB by development status (.e.g. alpha,beta,stable,mature,inactive) It is very common (in other langs at least) to have some approx development status. Otherwise it is really hard to choose. How can I know if a package is considered stable enough for production ? -- Regards, Gabi http://bugspy.net

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let expression
by John Moore 10 Feb '10

10 Feb '10

Hi All, Trying to get this to work, keeps telling me there a parse error on right in the let expression. Can anyone see where the problem is and be able to explain it to me. import Maybe data Expression = Val Double | Add Expression Expression | Subtract Expression Expression | Multiply Expression Expression | Divide Expression Expression | Var String | Let String Expression Expression deriving Show demo1 = (Add(Multiply(Divide(Subtract(Val 25)(Val 5))(Val 10))(Val 7))(Val 30)) type Dict =[(String,Expression)] emptyDict :: Dict emptyDict = [] addEntry :: String->Expression ->Dict -> Dict addEntry n e d = (n,e): d lookupEntry :: String -> Dict -> Maybe Expression lookupEntry n [] = Nothing lookupEntry n (x:xs) = if (n == k) then (Just v) else lookupEntry n xs where (k,v) = x evalStep :: Dict -> Expression -> Expression evalStep d(Val x)= (Val x) evalStep d(Add x y) = case x of (Val a) -> case y of (Val b) -> Val (a+b) left -> Add x (evalStep d y) right -> Add (evalStep d x)y evalStep d(Subtract x y) = case x of (Val a) -> case y of (Val b) -> Val (a-b) left -> Subtract x (evalStep d y) right -> Subtract (evalStep d x)y evalStep d(Multiply x y) = case x of (Val a) -> case y of (Val b) -> Val (a*b) left -> Multiply x (evalStep d y) right -> Multiply (evalStep d x)y evalStep d (Divide x y) = case x of (Val a) -> case y of (Val b) -> Val (a/b) left -> Divide x (evalStep d y) right -> Divide (evalStep d x)y evalStep d (Let n e1 e2) = case e1 of (Val a) -> case e2 of (Val b)-> Val (Let e1 e2) left -> Let e1 (evalStep d e2) right -> Let (evalStep d e1) e2 evaluate :: Dict-> [Expression] -> Expression -> IO() evaluate d(x:xs) e = do putStrLn (show e) putStrLn "Do another step (y/n) or rollback (r)? :" c <- getLine case c of "y" -> let e'= (evalStep d e)in evaluate d (e:x:xs) e'-- build up history "r" -> case (x:xs) of (x:xs)-> evaluate d xs x []-> do { putStrLn "Empty" ;evaluate d(x:xs) e } "n" -> putStrLn $ "Ok you said no :" ++ c John

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Re: Simplifying code
by edgar klerks 10 Feb '10

10 Feb '10

Hello all, I have a very interesting alternative solution of the problem. First I generate a tree with all the permutations: eg if I want to find al permutations of ao, I get the following tree [a @] | | [o 0] [o 0] Then I walk trough the tree so I can print it. There is only one ugly thing. Showtree' returns a string (in our example aoa0@o@0). To make it a list i put a \n between a left and right node and then use lines to make it a list of strings. Can someone point me in the right direction how to beautify it a bit? I also have an annoying problem with gmail and firefox. It seems it doesn add my posts to the current thread, but starts a new one. Oh well think I switch to evolution. I will review all yours solutions tomorrow. I saw some very beautifull things. With kind regards, Edgar Klerks module Main where import Data.Char data WordTree = Chain (Char,WordTree) | Choice (Char, WordTree) (Char, WordTree) | Stop instance Show WordTree where show = unlines.showTree type Rule = (Char, Char) type Rules = [Rule] infixl 4 ==> a ==> b = (a,b) rules :: Rules rules = [ 'a' ==> '@', 'l' ==> '|'] buildTree :: String -> Rules -> WordTree buildTree [] r = Stop buildTree (c:cs) r = case lookup c r of Just a -> Choice (a, buildTree cs r) (c, buildTree cs r) Nothing -> Chain (c, buildTree cs r) showTree a = lines $ showTree' a [] showTree' (Chain (a,b)) p = a : showTree' b p showTree' (Choice (a,b) (c,d)) p = c : showTree' d p ++ "\n" ++ (a : showTree' b p) showTree' (Stop) p = p

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Simplifying code
by edgar klerks 10 Feb '10

10 Feb '10

Hi All, I wrote a program which permutates a list words with some rules. For example if I feed it a list like: banana waterloo vraag It gives back the list: banana b@nana ban@na b@n@na banan@ b@nan@ ban@n@ b@n@n@ waterloo wa+erloo water|oo waterl0o etc However I have the feeling I am doing things to complicated. I am still a beginner. Would someone like to help me simplify somethings. If you think this is inappropriate please state also. I am not offended then. I understand you are offering your spare time to help me. The first thing I don't get is this. I recognize some things could be rewritten with a bind operator (because of the concat $ fmap), but I am puzzled how: mutateWords :: [Char] -> [[Char]] mutateWords word = nub.concat $ fmap snd <$> fmap unzip <$> ( foldr(\x z -> let char = snd x nm = number word lst = fst x in (insertAt char nm <$> lst) : z ) [[]] $ mw word ) Here is the full code: import Data.List import System import System.IO import Control.Applicative ---CONFIG section leat = ['s' ==> '$', 't' ==> '+', 'l' ==> '|', 'o' ==> '0','e' ==> '3', 'a' ==> '@', 'v' ==> '^'] leata = fst.unzip $ leat leatb = snd.unzip $ leat -- Perl like assoc lists infixl 1 ==> a ==> b = (a, b) -- Flipped fmap sometimes nicer infixl 4 <$$> xs <$$> f = f <$> xs -- first I need to find the positions of the mutatable charachters. findPositions :: [Char] -> [[Int]] findPositions xs = take (length index) $ index <*> [xs] where index = elemIndices <$> leata -- And generate all subsequences findSubSeq :: [Char] -> [[[Int]]] findSubSeq = fmap subsequences <$> findPositions -- Only change elements which needs to be changed insertAt :: Char -> [(Int, Char)] -> [Int] -> [(Int,Char)] insertAt c xs ps = xs <$$> (\x -> if (fst x) `elem` ps then (fst x , c) else x ) -- Couples character to mutable positions mw word = (findSubSeq word) `zip` leatb number = zip [0..] mutateWords :: [Char] -> [[Char]] mutateWords word = nub.concat $ fmap snd <$> fmap unzip <$> ( foldr(\x z -> let char = snd x nm = number word lst = fst x in (insertAt char nm <$> lst) : z ) [[]] $ mw word ) generateAll :: [Char] -> [[Char]] generateAll word = g lea $ mutateWords word where g 0 words = words g n words = g (n - 1) (nub $ words >>= mutateWords ) lea = length leata main = do filename <- getArgs wordlist <- readFile $ filename !! 0 let a = (words wordlist) >>= generateAll mapM_ putStrLn a -- Flatliner ICT Service, Email: Edgar.klerks(a)gmail.com, Tel: +31727851429 Fax: +31848363080 Skype: edgar.klerks Website: flatlinerict.nl Adres: Koelmalaan 258, 1813JD, Alkmaar Nederland

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Is there a an include statement equivalence in Haskell
by Andrew Smith 10 Feb '10

10 Feb '10

Hi I have a few lines of Haskell which I'd like to be pulled into my Haskell program, similar to the #include in C/C++, or Latex. I don't want to just paste the code permanently using editor, or pass it through say M4, nor build my own module. At the moment its a couple of lines I've used in three or four similar programs. As a Haskell beginner, is my thinking still stuck in the imperative coding style era ? -- Andrew

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