January 2008 - Haskell-Cafe

Type issues for a foldable trie in Haskell
by Brad Larsen 20 Jan '08

20 Jan '08

Hello there, I have written a trie in Haskell generalized to Eq a => [a] rather than simply String. I want to make this type an instance of Foldable, but I've run into a type dilemma. My datatypes look like this: data TrieElem a = Elem a | Start | End deriving (Read, Show, Eq, Ord) data Trie a = Trie {label :: TrieElem a ,children :: [Trie a]} deriving (Read, Show, Eq, Ord) The signature of Data.Foldable.foldr is (Data.Foldable.Foldable t) => (a -> b -> b) -> b -> t a -> b. However, I want the functions in Foldable to operate on the _list type_ that Trie stores rather than the _elements_ of that list type---Trie a stores lists of type a. For example, a Trie storing strings would have type Trie Char, and I want Trie Char to be Foldable, but where the functions operate on String rather than Char. So, with the datatype definitions of Trie and TrieElem as they are above, to define a fold function that operates the way I want would have signature ([a] -> b -> b) -> b -> Trie a -> b, which is no good for making Trie a an instance of Foldable. Hopefully this doesn't just seem like rambling :-). How might I rewrite my datatypes to do what I want, preferably without using ghc extensions? Thanks! Brad Larsen

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Basic binary IO
by Jamie Love 20 Jan '08

20 Jan '08

Hello all, I'm wondering if anyone has a reference to any binary IO and data conversion tutorials. I'm playing around with generating a BMP file in haskell, and am a little stuck on the "best" way to go about the simple task of creating the BMP header. The header is "BM" + 4 bytes for file size + 4 bytes reserved + 4 bytes for offset where data begins. I have the basis starting off at: bmpHeader = B.pack $ [ 0x42, 0x4D ] ++ [0 , 0, 0, 0] ++ [0 , 0, 0, 0] ++ [14 :: Int32] (where B is Data.ByteString) I'm wondering how I can: 1/ convert a 32 bit number (Int32, Char32) to 4 Char8 elements 2/ rotate bits/bytes in a 32 bit Char32 (or Int32) so they are explicitly little-endian (I work on a mac powerbook, and it is big-endian) 3/ convert an Integer or Int type to an Int32 type Any pointers or suggestions would be helpful. Thanks -- Jamie Love Senior Consultant Aviarc Australia Mobile: +61 400 548 048 ------------------------------------------------------------ This message has been scanned for viruses and dangerous content by MailScanner and is believed to be clean.

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One-line haskell program with GHC
by Sukit Tretriluxana 19 Jan '08

19 Jan '08

Hi, I don't know if it's been asked before. I just wonder if GHC supports some sort of one-liner program that can be specify right as the argument to either ghci or runghc program. In perl, they have something like perl *-e* 'print "Hello"' Do we have similar thing with GHC? Thanks, Ed

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overlapping instance error -- need help using instance Monoid (Sum a) where a is a map of "money" values
by Thomas Hartman 19 Jan '08

19 Jan '08

The code below compiles as given, however if I uncomment the tSM function I get the overlapping instance error mentioned in the subject line. Can someone give me advice on how to do what I want to do? basically I want to add, for example, (USD,1) and (USD,2) and (Euro,3) and get result fromList [(USD,3), (Euro,3)] the datatypes below are more verbose but the above is the basic idea thanks! thomas. ***** module TransactionRows {- ( mkTransactionRows,TransactionRows ) -} where import Data.Monoid import Data.List import qualified Data.Map as M data Currency a = Currency a deriving ( Show, Eq, Ord ) data Money c a = Money ( M.Map c a ) deriving Show instance (Num a, Ord c)=> Monoid ( Sum (Money c a) ) where mempty = Sum ( Money M.empty ) Sum ( Money m1 ) `mappend` Sum ( Money m2 ) = Sum (Money m1 `plusmoney` Money m2) plusmoney (Money m1s) (Money m2s) = Money msum where msum = foldl' f m1s (M.toList m2s) f m (k,v) = M.insertWith (+) k v m mkMoney1 :: Currency String -> Float -> Money (Currency String) Float mkMoney1 c a = Money $ M.singleton c a --sumMoney :: [Money (Currency String) Float] -> Money (Currency String) Float sumMoney ms = getSum $ mconcat $ map Sum ms -- if I uncomment this, get Overlapping instances for Monoid error --tSM = sumMoney [mkMoney1 (Currency "usd") 1 :: Money (Currency String) Float]

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Re: [Haskell-cafe] Re: 0/0 > 1 == False
by Jonathan Cast 19 Jan '08

19 Jan '08

On 19 Jan 2008, at 2:52 AM, Kalman Noel wrote: > Jonathan Cast wrote: >> On 12 Jan 2008, at 3:23 AM, Kalman Noel wrote: >>> (2) lim a_n = ∞ > [...] >>> (2) means that the sequence does not converge, >> >> To a value in R. Again, inf is a perfectly well defined extended >> real number, and behaves like any other element of R u {-inf, inf}. >> (Although that structure isn't quite a field --- 0 * inf isn't >> defined, nor is inf - inf). > > Out of curiosity, is there some typical application domain for > extended real > numbers? In calculus and elementary analysis, the notion of limits at/to infinity, improper Riemann integrals, etc., are introduced by a succession of `special notations'. Taking the extended real numbers as the underlying space permits these notations to be defined more compositionally, because ∞ is now an ordinary mathematical object. For example, if f is a nonnegative measurable function, ∫f on a measurable set is /always/ defined (as an extended real number) and the special case of an `integrable' function is simply one where the integral (which is an actual mathematical value) is an element of R. So, when we say ∫f ∈ R, that notation is compositional --- that's real set membership there. Similarly, ∫_{-∞}^∞ f is defined (as a Lebesgue integral) the same way any other integral is, because the interval [-∞, ∞] is a perfectly good mathematical object. jcc

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0/0 > 1 == False
by Mitar 19 Jan '08

19 Jan '08

Hi! Why is 0/0 (which is NaN) > 1 == False and at the same time 0/0 < 1 == False. This means that 0/0 == 1? No, because also 0/0 == 1 == False. I understand that proper mathematical behavior would be that as 0/0 is mathematically undefined that 0/0 cannot be even compared to 1. There is probably an implementation reason behind it, but do we really want such "hidden" behavior? Would not it be better to throw some kind of an error? Mitar

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functional graphs
by Thomas Hartman 19 Jan '08

19 Jan '08

I don't get where the graph edges (Node,Node) are specified. Do you just assume that every two nodes have an edge between them and calculate the edge label from the function? Also, is there a real world / motivating use for a graph defined this way? thomas. 2008/1/18, Christian Maeder <Christian.Maeder(a)dfki.de>: > Hi, > > Given a complete graph as a list of nodes whose edge labels are given by > a function over two nodes: > > data CGraph a b = CGraph [a] (a -> a -> b) > > Can I define an instance for the fgl Graph class? > > import Data.Graph.Inductive.Graph > > instance Graph CGraph where > empty = CGraph [] -- and now? > > I had no idea how to define empty (except using undefined). > > I thought of requiring a context for the node labels of type a, but this > type is not mentioned in the class header. So it looked to me like the > impossibility to define sets (requiring an Ord) as monads. (i.e. > instance Monad Data.Set.Set) > > Any working proposals for my graph problem? > > Cheers Christian > _______________________________________________ > Haskell-Cafe mailing list > Haskell-Cafe(a)haskell.org > http://www.haskell.org/mailman/listinfo/haskell-cafe >

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First go at reactive programming
by Levi Stephen 19 Jan '08

19 Jan '08

Hi, Listed below is my first experiment with reactive programming. It is a simple web server written using the Data.Reactive[1] library. The intended interface is given by the runHttpServer function, so the remainder is intended to be internal. I'd be happy to hear comments on any parts of this, but am particularly interested in the following: 1. Is this kind of code what is intended from reactive programming? 2a. I'm not sure about passing the (Handle,...) tuple around. Is there a way to avoid this? 2b. I'm not sure of the best place to handle possible socket exceptions 2c. I'd like to be able to pass a function of type Event Request -> Event Response to runHttpServer, so that reactive programming could be used throughout client code also, but the (Handle,...) tuples seem to be getting in the way. 3. I have a feeling there's a clearer way to write responseSend. Thanks, Levi [1] http://www.haskell.org/haskellwiki/Reactive module Main where import Control.Applicative import Control.Arrow ((&&&),(>>>)) import Control.Concurrent import Control.Monad import Data.Reactive import Network.BSD import Network.HTTP import Network import System.IO import Text.XHtml.Strict type RequestHandler = Request -> Response main = runHttpServer helloWorldHandler helloWorldHandler :: RequestHandler helloWorldHandler _ = Response (2,0,0) "" [] $ prettyHtml helloWorldDoc helloWorldDoc = header << thetitle << "Hello World" +++ body << h1 << "Hello World" runHttpServer r = socketServer >>= runE . handleConnection r socketServer :: IO (Event Handle) socketServer = withSocketsDo $ do (e,snk) <- mkEventShow "Server" sock <- listenOn (PortNumber 8080) forkIO $ forever $ acceptConnection sock $ snk return e handleConnection :: RequestHandler -> Event Handle -> Event (IO ()) handleConnection r = handleToRequest >>> runRequestHandler r >>> responseSend handleToRequest :: Event Handle -> Event (Handle, IO (Result Request)) handleToRequest e = fmap (id &&& receiveHTTP) e responseSend :: Event (Handle, IO (Result Response)) -> Event (IO ()) responseSend e = fmap (\(h,rsp) -> rsp >>= either (putStrLn . show) (respondHTTP h) >> close h) e runRequestHandler :: RequestHandler -> Event (Handle, IO (Result Request)) -> Event (Handle, IO (Result Response)) runRequestHandler r e = fmap hrToHr e where rqhdl :: Result Request -> Result Response rqhdl rq = bindE rq (Right . r) hrToHr :: (Handle, IO (Result Request)) -> (Handle, IO (Result Response)) hrToHr (h,req) = (h, liftA rqhdl req) acceptConnection s k = accept s >>= \(h,_,_) -> forkIO $ k h instance Stream Handle where readLine h = hGetLine h >>= \l -> return $ Right $ l ++ "\n" readBlock h n = replicateM n (hGetChar h) >>= return . Right writeBlock h s = mapM_ (hPutChar h) s >>= return . Right close = hClose

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Haskell-Support in Ohloh
by Joachim Breitner 19 Jan '08

19 Jan '08

Hi, I’m only margianlly involved or up to date there, but still some might have missed this: Ohloh has begun to release their tools, starting with ohcount, their tool to measure code and comments lines: http://labs.ohloh.net/ohcount/ They explicitly write that they want haskell support, and the oldest open bug report on their page is about this: http://labs.ohloh.net/ohcount/ticket/205 So if anyone feels like programming some ruby (I guess they want it to be in that language as well) and wants to give the haskell community a chance for wider audience, give it a shot. Greetings, Joachim -- Joachim "nomeata" Breitner mail: mail(a)joachim-breitner.de | ICQ# 74513189 | GPG-Key: 4743206C JID: nomeata(a)joachim-breitner.de | http://www.joachim-breitner.de/ Debian Developer: nomeata(a)debian.org

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How best to make GHC 6.6 and 6.8 co-exist on one server/account
by Steve Lihn 19 Jan '08

19 Jan '08

Hi, It appears some of the latest hackages are moving towards 6.8 to take advantage of the new features, while quite a few remains at 6.6. The compatibility between the two versions has been problematic. I only have 6.6 installed, but now thinking to add 6.8 to my account (so I can switch between them, depending what works where). What is the best practice to make 6.8 co-exist with 6.6 ? Specific questions are: 1) I installed 6.6 under prefix=$HOME/ghc. If I install 6.8 to the same prefix, it will overwrite 6.6. Correct? How do I get around it? 2) I installed additional tools and packages under $HOME/htools using 6.6. (Which I begin to wonder if that is necessary.) Do I have to use a different directory for 6.8? I noticed that, for each package, the path is $HOME/htools/lib/<package-ver>/ghc-6.6.1/ etc etc. It seems that 6.8 will create subdirectories, so they can co-exist peacefully. Correct? Thanks, Steve

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