June 2009 - Haskell-Cafe

How to improve below code?
by Andy Stewart 09 Jun '09

09 Jun '09

Hi all, I have below source code, i use Dynamic for `pageBuffer`. In implement of function `pageClone`, after `case pt of`, i need write like this, and this code looks ugly, if `PageTyep` have 100 type, i need write those ugly code 100 times. case pt of TStringBuffer -> pageBufferClone (x :: StringBuffer) TImageBuffer -> pageBufferClone (x :: ImageBuffer) TVideoBuffer -> pageBufferClone (x :: TVideoBuffer) TMixBuffer -> pageBufferClone (x :: TMixBuffer) So have a better solution to avoid write above ugly code? Someone suggestion me use GADTs instead, but i don't know how to write correct GADTs code replace current version, if GADTs is best way, someone can explain it detail? It's better if someone give me demo code. ------------------------------> Source Code start <------------------------------ data PageType = TStringBuffer | TImageBuffer | TVideoBuffer | TMixBuffer deriving (Eq, Show, Read) data Page = Page {pageName :: IORef String ,pageId :: Int ,pageType :: PageType ,pageBuffer :: Dynamic ,pageBox :: VBox } class PageBuffer a where pageBufferClone :: a -> IO (Dynamic, VBox) -- | Page clone interface. pageClone :: Page -> IO Page pageClone page = do -- Get common information for clone page. name <- pageGetName page let id = pageId page pt = pageType page pb = pageBuffer page -- Get clone information for dynamic interface. (pBuffer, pBox) <- case fromDynamic pb of Just x -> case pt of TStringBuffer -> pageBufferClone (x :: StringBuffer) TImageBuffer -> pageBufferClone (x :: ImageBuffer) TVideoBuffer -> pageBufferClone (x :: TVideoBuffer) TMixBuffer -> pageBufferClone (x :: TMixBuffer) Nothing -> pageCloneEmpty -- Return clone page. pageNewInternal name id pt pBuffer pBox ------------------------------> Source Code end <------------------------------ Thanks! -- Andy

3 2

please comment on my parser, can I do this cleaner?
by Thomas Hartman 09 Jun '09

09 Jun '09

All I want to do is split on commas, but not the commas inside () or <> tags. I have been wanting to master parsec for a long time and this simple exercise looked like a good place to start. The code below does the right thing. Am I missing any tricks to make this simpler/neater? Thanks, thomas. thartman@ubuntu:~/perlArena>cat splitEm. splitEm.hs splitEm.hs~ splitEm.pl splitEm.pl~ thartman@ubuntu:~/perlArena>cat splitEm.hs {-# LANGUAGE ScopedTypeVariables #-} import Text.ParserCombinators.Parsec import Text.ParserCombinators.Parsec.Char import Text.PrettyPrint (vcat, render, text) import Data.List.Split hiding (sepBy, chunk) import Text.ParserCombinators.Parsec.Token import Debug.Trace import Debug.Trace.Helpers -- this works, but is there a neater/cleaner way? main = ripInputsXs (toEof splitter) "splitter" [ goodS, badS ] -- I need a way to split on commas, but not the commas inside '<>' or '()' characters goodS = "<*2>FOO<2,1>,<*3>(SigB<8:0:2>,BAR),<*2>Siga<2:0>,Sigb<8,7,6,5,0>" badS = "<*2)FOO<2,1>,<*3>(SigB<8:0:2>,BAR),<*2>Siga<2:0>,Sigb<8,7,6,5,0>" -- the first < matches a ), so reject this splitter = do chunks :: [String] <- toEof (many chunk) let pieces = map concat $ splitOn [","] chunks return pieces -- chunks where atom = string "," <|> ( many1 $ noneOf "()<>" ) chunk = parenExpr <|> atom parenExpr :: GenParser Char st [Char] parenExpr = let paren p = betweenInc (char '(' ) (char ')' ) p <|> betweenInc (char '<' ) (char '>' ) p in paren $ option "" $ do ps <- many1 $ parenExpr <|> atom return . concat $ ps betweenInc o' c' p' = do o <- o' p <- p' c <- c' return $ [o] ++ p ++ [c] toEof p' = do r <- p' eof return r ripInputs prs prsName xs = mapM_ (putStrLn . show . parse prs prsName ) xs ripInputsXs prs prsName xs = mapM_ (putStrLn . showXs . parse prs prsName ) xs where showXs v = case v of Left e -> show e Right xs -> render . vcat . map text $ xs

2 1

Re: [Haskell-cafe] Applying Data.Map
by michael rice 09 Jun '09

09 Jun '09

In the import statements, it wasn't clear to me that I could import types as well as functions, and Map is a type. All clear now. Thanks. Michael --- On Tue, 6/9/09, Thomas ten Cate <ttencate(a)gmail.com> wrote: From: Thomas ten Cate <ttencate(a)gmail.com> Subject: Re: [Haskell-cafe] Applying Data.Map To: "michael rice" <nowgate(a)yahoo.com> Cc: haskell-cafe(a)haskell.org Date: Tuesday, June 9, 2009, 9:40 AM On Tue, Jun 9, 2009 at 15:23, michael rice<nowgate(a)yahoo.com> wrote: > import Data.Map (Map) (fromList,!) ??? > import qualified Data.Map as Map (fromList,!) ??? Because ! is an operator, you need to enclose it in parentheses. Also, the (Map) in the import is already the list of things you are importing; you can just add to that. So do the following: Import these without qualification: > import Data.Map (Map, fromList, (!)) Import everything else (actually including Map, fromList and (!)) with qualification "Map": > import qualified Data.Map as Map Cheers, Thomas

1 0

Re: [Haskell-cafe] Applying Data.Map
by michael rice 09 Jun '09

09 Jun '09

Hi Toby, Thanks for the helpful comments. I'd gotten used to arithmetic operator sections (+2), (*2), etc. but hadn't picked up on the generality of using them with *any* infix function. I can also see the benefit of using List.Group. However, I'm uncertain about how to import just fromList and ! from with the imports I'm using import Data.Map (Map) (fromList,!) ??? import qualified Data.Map as Map (fromList,!) ??? Michael --- On Mon, 6/8/09, Toby Hutton <toby.hutton(a)gmail.com> wrote: From: Toby Hutton <toby.hutton(a)gmail.com> Subject: Re: [Haskell-cafe] Applying Data.Map To: "michael rice" <nowgate(a)yahoo.com> Cc: haskell-cafe(a)haskell.org Date: Monday, June 8, 2009, 8:57 PM Although in this example using Data.Map is overkill, if the alphabet was very large then Data.Map probably would be the way to go. In that case I'd use: map head . group . sort instead of nub . sort since it's noticeably quicker for large lists. This is because nub needs to preserve the order of input, removing redundancies, but you're sorting it anyway. Also, in map (\c -> m Map.! c) s you can use the 'section' (m Map.!) instead. e.g., map (m Map.!) s The Map.! is ugly though. As you're only using fromList and (!) from Data.Map, I'd just import those explicitly since they don't clash with Prelude. Then you'd have map (m !) s Toby. On Tue, Jun 9, 2009 at 4:59 AM, michael rice <nowgate(a)yahoo.com> wrote: I wrote a Haskell solution for the Prolog problem stated below. I had written a function SQUISH before discovering that NUB does the same thing. While the solution works, I thought maybe I could apply some functions in the Data.Map module, and so wrote a second version of SERIALIZE, one no longer needing TRANSLATE. Using the Data.Map module is probably overkill for this particular problem, but wanted to familiarize myself with Map type. Suggestions welcome. Prolog code also included below for those interested. Michael =========== {- From "Prolog By Example", Coelho, Cotta, Problem 42, pg. 63 Verbal statement: Generate a list of serial numbers for the items of a given list, the members of which are to be numbered in alphabetical order. For example, the list [p,r,o,l,o,g] must generate [4,5,3,2,3,1] -} {- Prelude> :l serialize [1 of 1] Compiling Main ( serialize.hs, interpreted ) Ok, modules loaded: Main. *Main> serialize "prolog" [4,5,3,2,3,1] *Main> -} ===========Haskell code========== import Data.Char import Data.List import Data.Map (Map) import qualified Data.Map as Map {- translate :: [Char] -> [(Char,Int)] -> [Int] translate [] _ = [] translate (x:xs) m = (fromJust (lookup x m)) : (translate xs m ) -} {- serialize :: [Char] -> [Int] serialize s = let c = nub $ sort s n = [1..(length c)] in translate s (zip c n) -} serialize :: [Char] -> [Int] serialize s = let c = nub $ sort s n = [1..(length c)] m = Map.fromList $ zip c n in map (\c -> m Map.! c) s ============Prolog code============ serialize(L,R) :- pairlists(L,R,A),arrange(A,T), numbered(T,1,N). ? <- typo? pairlists([X|L],[Y|R],[pair(X,Y)|A]) :- pairlist(L,R,A). pairlists([],[],[]). arrange([X|L],tree(T1,X,T2)) :- partition(L,X,L1,L2), arrange(L1,T1), arrange(L2,T2). arrange([],_). partition([X|L],X,L1,L2) :- partition(L,X,L1,L2). partition([X|L],Y,[X|L1],L2) :- before(X,Y), partition(L,Y,L1,L2). partition([X|L],Y,L1,[X|L2]) :- before(Y,X), partition(L,Y,L1,L2). partition([],_,[],[]). before(pair(X1,Y1),pair(X2,Y2)) :- X1<X2. numbered(tree(T1,pair(X,N1),T2),N0,N) :- numbered(T1,N0,N1), N2 is N1+1, numbered(T2,N2,N). numbered(void,N,N). ============Prolog examples======== Execution: ?- serialize([p,r,o,l,o,g]). [4,5,3,2,3,1] ?- serialize ([i,n,t,.,a,r,t,i,f,i,c,i,a,l]). [5,7,9,1,2,8,9,5,4,5,3,5,2,6] _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe(a)haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe

2 1

Combine to List to a new List
by ptrash 09 Jun '09

09 Jun '09

Hi, I have the following two lists: a = [1,2,3] b = ["A","B","C"] I want a combination of the to lists: c = [(1,"A"), (2, "B"), (3, "C")] How can I do this? I have tried c = [(x,y) | x <- a, y <- b] But this just returns me a list with every possible combination of the 2 lists. Thanks... -- View this message in context: http://www.nabble.com/Combine-to-List-to-a-new-List-tp23942440p23942440.html Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com.

4 4

A generics question
by Henry Laxen 09 Jun '09

09 Jun '09

Lets suppose I have a file that has encoded things of different types as integers, and now I would like to convert them back into specific instances of a data type. For example, I have a file that contains 1,1,2,3 and I would like the output to be [Red, Red, Green, Blue]. I also would like to do this generically, so that if I wanted to convert the same list of integers into say Sizes, I would get [Small, Small, Medium, Large] Now please have a look at the following code: {-# LANGUAGE DeriveDataTypeable #-} import Data.Generics data Color = Red | Green | Blue deriving (Eq,Ord,Read,Show,Typeable,Data) data Size = Small | Mediaum | Large deriving (Eq,Ord,Read,Show,Typeable,Data) g = Green c = undefined :: Color s = undefined :: Size t = do print $ toConstr g -- Green print $ dataTypeOf c -- DataType {tycon = "Main.Color", datarep = AlgRep [Red,Green,Blue]} convert :: (Data a, Data b) =Int -a -b convert i x = let c = dataTypeConstrs (dataTypeOf x) !! (i-1) in fromConstr c I would like to be able to say: x = convert 1 c and have it assign Red to x then I would like to say: y = convert 1 s and have it assign Small to y, however, when I try that I get: Ambiguous type variable `b' in the constraint: `Data b' arising from a use of `convert' at <interactive>:1:8-18 Probable fix: add a type signature that fixes these type variable(s) Of course if I say x :: Color = convert 1 c, it works, but I would like to avoid that if possible, as all of the information is already contained in the parameter c. Is there any way to do this? Thanks in advance for your wise counsel. Best wishes, Henry Laxen

4 5

Unification for rank-N types
by Vladimir Reshetnikov 09 Jun '09

09 Jun '09

Hi, I have the following code: ------------------------------------------------------- {-# LANGUAGE RankNTypes #-} f :: ((forall a. a -> a) -> b) -> b f x = x id g :: (forall c. Eq c => [c] -> [c]) -> ([Bool],[Int]) g y = (y [True], y [1]) h :: ([Bool],[Int]) h = f g ------------------------------------------------------- GHC rejects it: Couldn't match expected type `forall a. a -> a' against inferred type `forall c. (Eq c) => [c] -> [c]' Expected type: forall a. a -> a Inferred type: forall c. (Eq c) => [c] -> [c] In the first argument of `f', namely `g' In the expression: f g But, intuitively, this code is type-safe, and actually I can convince the typechecker in it with the following workaround: ------------------------------------------------------- h :: ([Bool],[Int]) h = let g' = (\(x :: forall a. a -> a) -> g x) in f g' ------------------------------------------------------- So, is the current behavior of GHC correct ot it is a bug? How unification for rank-N types should proceed? Thanks, Vladimir

1 1

Dutch HUG meeting tonight in Amsterdam
by Chris Eidhof 09 Jun '09

09 Jun '09

Hi everyone, Tonight there will be another meeting of the Dutch Haskell Users' Group! This time we'll meet in Amsterdam, in the library. On the wiki [1] you can find the details of how to reach it. We'll be at the top floor and shouldn't be hard to recognize. The meeting is set to begin at 19:30. Everybody's welcome, from beginners to advanced haskellers. Even if you never programmed in Haskell before it'll probably be a lot of fun. See you tonight! There are a lot of international people joining, so no knowledge of Dutch is necessary. If you're joining, also be sure to subscribe to our mailinglist [2]. See you tonight! -chris [1]: http://www.haskell.org/haskellwiki/Dutch_HUG [2]: http://groups.google.com/group/dutch-hug

1 0

How to compile base?
by Henk-Jan van Tuyl 09 Jun '09

09 Jun '09

L.S., I tried to compile base-4.0.0.0 (on Windows XP) as follows: [...]\base\4.0.0.0>runhaskell Setup configure <command line>: module `Prelude' is not loaded It seems that Base needs another way to compile, how? -- Regards, Henk-Jan van Tuyl -- http://functor.bamikanarie.com http://Van.Tuyl.eu/ --

2 1

Deprecated packages on Hackage?
by Erik de Castro Lopo 09 Jun '09

09 Jun '09

Hi all, Is there a way to list all the deprecated packages on hackage? For instance I found this: http://hackage.haskell.org/cgi-bin/hackage-scripts/package/control-timeout-… via a google search but its not listed on the main hackage packages page. However, there are other packages which are marked DEPRECATED in their description strings. Finally, if a package is deprecated it might be usefult to have a reason as well so the hackage entry might say: Deprecated : true (replaced by package XXX) or Deprecated : true (needs maintainer) Cheers, Erik -- ---------------------------------------------------------------------- Erik de Castro Lopo http://www.mega-nerd.com/

3 3