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Data.Binary Endianness
by Adrian Neumann 16 Sep '07

16 Sep '07

-----BEGIN PGP SIGNED MESSAGE----- Hash: RIPEMD160 I try to write a Windows Bitmap File using Data.Binary, but I have some trouble. For example, the internet states, that the magic number, that puts 'BM' in the first two bytes of the file is 19778. But when I > put (19778::Word16) I get 'MB' instead. I read on the german Wikipedia, that bmp uses little endian encoding, but Data.Binary uses big endian. Obviously this can't work. Is there a way to switch endianness? Or at least a function that converts between the two? I looked at the Data.Binary Haddock documentation, but found nothing like that. Thanks in advance, Adrian -----BEGIN PGP SIGNATURE----- Version: GnuPG v1.4.7 (MingW32) Comment: Using GnuPG with Mozilla - http://enigmail.mozdev.org iD8DBQFG5Q1Z11V8mqIQMRsRA/V6AJoC/R2k5rz0LjQI4iGoi79zLYZmgwCdEx/g cZOR5jseUcQnc6+I09N+xj8= =K5ww -----END PGP SIGNATURE-----

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unknown RTS option: -N2
by Gregory Propf 16 Sep '07

16 Sep '07

I've built a program with the -threaded option using ghc. This option is supposed to link your program to the threaded runtime with support for multicore CPUS (mine is a dual core). The program pukes with the message in the subject line when I try to use the -N option to tell it to use both CPUs. I'm reasonably sure that I actually have the threaded runtime because initially I got a runtime message from the program about how GTK is single threaded and can only be used with the threaded runtime using runtimeunsafeInitGUIForThreadedRTS instead of the standard initGUI. I made the change in the code and now the program runs but still does not know the -N RTS option. This problem seems to affect all my programs. It is as if the threaded runtime is there but doesn't know any of its special flags. This is the line I'm compiling the program with. It's a life game clone in case you're curious. Compiler is ghc-6.4.2 on 64 bit AMD dual core running Gentoo Linux. - Greg ghc -threaded -package gtk -package glade -o hlife hlgtk.o Hlife.o ____________________________________________________________________________________ Check out the hottest 2008 models today at Yahoo! Autos. http://autos.yahoo.com/new_cars.html

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Re: [Haskell] question about a failure to generalize
by Tim Chevalier 16 Sep '07

16 Sep '07

[replying to haskell-cafe] On 9/15/07, Norman Ramsey <nr(a)eecs.harvard.edu> wrote: > Dear Haskellers, > > I've had a Haskell program rejected for reasons I don't understand. > Here's the relevant bit of code; the problem is that I expected the > type-inference engine to generalize the abbreviation 'fold' to an > overloaded function, but it doesn't---to make the code work, I had to > expand 'fold' into 'foldRegsUsed' everywhere it appears. I'm baffled > because 'fold' isn't mutually recursive with anything, so I would have > expected ordinary Hindley-Milner style inference to generalize it to > something of the type > > UserOfLocalRegs a => (b -> LocalReg -> b) -> b -> a -> b > > But that's not what happens. This code fails to compile because the > compiler is willing to use 'fold' at only one type (CmmExpr as it happens): > > class UserOfLocalRegs a where > foldRegsUsed :: (b -> LocalReg -> b) -> b -> a -> b > > instance UserOfLocalRegs Middle where > foldRegsUsed f z m = middle m > where middle (MidComment {}) = z > middle (MidAssign _lhs expr) = fold f z expr > middle (MidStore addr rval) = fold f (fold f z addr) rval > middle (MidUnsafeCall tgt _ress args) = fold f (fold f z tgt) args > middle (CopyIn _ _formals _) = z > middle (CopyOut _ actuals) = fold f z actuals > fold = foldRegsUsed > > > What rule of the language have I overlooked? > Monomorphism restriction? Replacing fold with foldRegsUsed would work because there's a type signature for foldRegsUsed. Cheers, Tim -- Tim Chevalier * catamorphism.org * Often in error, never in doubt "...There is no mystery; there is only paradox, the incontrovertible union of contradictory truths." -- Edward Abbey

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Sequencing Operations in a Monad
by Dominic Steinitz 15 Sep '07

15 Sep '07

> If you arrange the types to try to do all the operations inside the IO > monad you can't chain together more than 1 binary operation. eg. > > do > S <- A + B > Z <- Q * S > > vs > > do > S <- Q * (A + B) > > Are there any suggestions for this dilemma? Am I using the wrong monad for > this task? I'm not sure if this is what you are asking but isn't liftM2 or some variant what you need? Dominic.

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haskell on llvm?
by brad clawsie 15 Sep '07

15 Sep '07

has anyone ever considered using llvm as a infrastructure for haskell compilation? it wold seem people are looking at building frontends for scheme, ocaml, etc. i don't know if an alternate backend is appropriate, but it would seem to be an interesting way to aggregate the best thinking for various optimizations over a more diverse group of developers.

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Is "take" behaving correctly?
by PR Stanley 15 Sep '07

15 Sep '07

Hi take 1000 [1..3] still yields [1,2,3] I thought it was supposed to return an error. Any ideas? Thanks, Paul

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Memory leak or wrong use of Array ?
by L.Guo 15 Sep '07

15 Sep '07

Hi MailList Haskell-Cafe: I am tring to solve Project Euler problem 70. And write some code. (will at the end of this mail) And, I run the code in GHCi. The problem is that, when the input is 1,000,000, it works fine, when the input is up to 10,000,000, the memory GHCi used increase very fast and did not stop. Is this a memory leak ? or, is there some mis-understand about array ? Regards -------------- -- Mudules : import Data.Array.IO import Foreign ( unsafePerformIO ) -- Codes : p070_solve = putStrLn . show $ solutionOf 10000000 where isPerm a b = sort (show a) == sort (show b) phis n = unsafePerformIO $ do arr <- newArray (2,n) (False,1/1) :: Fractional t => IO (IOArray Int (Bool,t)) mapM_ (sieve arr n) [2..n] factors <- getElems arr return . map (\(n,(b,f)) -> (n,floor $ toRational n*f)) $ zip [2..n] factors where sieve arr ubound p = do (b,o) <- readArray arr p if b then return () else mapM_ (update arr (toRational p)) . takeWhile (<=ubound) $ iterate (+p) p update arr p i = do (_,o) <- readArray arr i writeArray arr i (True,o*(p-1)/p) solutionOf = snd . minimum . map (\(n,phi)->(toRational n / toRational phi,n)) . filter (uncurry isPerm) . phis -------------- L.Guo 2007-09-14

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ANN: HPDF 1.1
by alpheccar 14 Sep '07

14 Sep '07

I have just released the version 1.1 of HPDF. You can get it on hackage: http://hackage.haskell.org/cgi-bin/hackage-scripts/package/HPDF-1.1 In this release, I am experimenting with typesetting. -- Christophe.

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Clarification Please
by PR Stanley 14 Sep '07

14 Sep '07

Hi Taken from chapter 6, section 8 of the Hutton book on programming in Haskell: 5. Using merge, define a recursive function msort :: (Ord a) => [a] -> [a] that implements merge sort, in which the empty list and singleton lists are already sorted, and any other list is sorted by merging together the two lists that result from sorting the two halves of the list separately. : Hint: first define a function ¬halve :: [a] -> [([a], [a])] ¬that splits a list into two halves whose length differs by at most one. Create a halve function - okay, that's fairly straightforward. The rest, I'm afraid, is a little obscure. I'm not looking for the solution; I'd like to work that out for meself. However, I'd really appreciate some clues as to the general structure of the algorithm. Much obliged, Paul

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getting crazy with character encoding
by Andrea Rossato 14 Sep '07

14 Sep '07

Hi, supposed that, in a Linux system, in an utf-8 locale, you create a file with non ascii characters. For instance: touch abèèè Now, I would expect that the output of a shell command such as "ls ab*" would be a string/list of 5 chars. Instead I find it to be a list of 8 chars...;-) That is to say, each non ascii character is read as 2 characters, as if the string were an ISO-8859-1 string - the string is actually treated as an ISO-8859-1 string. But when I print it, now it is displayed correctly. I don't understand what's wrong and, this is worse, I don't understand what I should be studying to understand what I'm doing wrong. After reading about character encoding, the way the linux kernel manages file names, I would expect that a file name set in an utf-8 locale should be read by locale aware application as an utf-8 string, and each character a unicode code point which can be represented by a Haskell char. What's wrong with that? Thanks for your kind attention. Andrea Here the code to test my problem. Before creating the file remember to set the LANG environmental variable. Something like: export LANG="en_US.utf8" should be fine. (Check your available locales with "locale -a") import System.Process import System.IO import Control.Monad main = do l <- fmap lines $ runProcessWithInput "/bin/bash" [] "ls ab*" putStrLn (show l) mapM_ putStrLn l mapM_ (putStrLn . show . length) l runProcessWithInput cmd args input = do (pin, pout, perr, ph) <- runInteractiveProcess cmd args Nothing Nothing hPutStr pin input hClose pin output <- hGetContents pout when (output==output) $ return () hClose pout hClose perr waitForProcess ph return output

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