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Adding to a package
by AbdulSattar Mohammed 21 Feb '12

21 Feb '12

I just submitted a patch to Wai, but I didn't like the process I went through while testing it on my machine. I cloned the repo, modified it and ran runghc Setup configure --prefix=$HOME --user runghc Setup install That registered as something like wai-2.1.1.1. ( I don't know what it exactly looked like, but it appended a ".1" to the previous version.) When I ran my test code, it used the latest version. Fine. But, is this the way everyone does that? I mean, adding a new version? I can't think of any other way, but I'm just against registering the package into the GHC registry. How does everyone do that? -- Warm Regards, AbdulSattar Mohammed

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A question about import
by bahadýr altan 19 Feb '12

19 Feb '12

Hi, when I try to use import like these: import System.Random import Control.Monad I get this error: "Syntax error in input (unexpected keyword "import")". Can you tell me why this happens and how I can fix it? Thanks..

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A simple remove function
by bahadýr altan 19 Feb '12

19 Feb '12

Hello. I'm trying to write a code which deletes the wanted element in a list. Like this : myremove "doaltan" 3 = "doltan". I wrote the code below but it doesn't even compile. Could you help me with fixing the code? Thanks! myremove x s = (take (s-1) x):(drop (s+1) x)

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A question about pattern matching
by bahadýr altan 18 Feb '12

18 Feb '12

Hello. I'm wondering what " (Eq a) =>" part in the code below does.. I couldn't figure it myself.. Thanks in advance :) isPalindrome :: (Eq a) => [a] -> Bool isPalindrome xs = xs == (reverse xs)

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The dot operator.
by bahadýr altan 18 Feb '12

18 Feb '12

Hello. I saw the dot "." operator in some Haskell codes, but I couldn't figure out what it does. I didn't see it in some books and tutorials and couldn't find an explanation on the Internet. Could you tell me what it does? Thanks:)

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my own last function
by bahadýr altan 18 Feb '12

18 Feb '12

Hello everyone, I have just started studying and started trying to solve 99 questions in haskell.org. However I got stuck even in the first question. The first question wants me to write my own last function and I wrote this : q1mylast :: [x] -> x q1mylast [a] = if length [a] <=1 then a else q1mylast (tail [a]) This program doesn't work and it gives this error : Program error: pattern match failure: q1mylast "doaltan" I'll be very glad if you could tell me what's wrong with the code and how can I fix it.. Thanks in advance :)

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Cabal complains that cabal-dev “indirectly depends on multiple versions of the same package”, even though it (apparently) does not
by James Fisher 16 Feb '12

16 Feb '12

Hi all, I've just asked this on SO as well ( http://stackoverflow.com/questions/9301122/haskell-cabal-package-indirectly…) but to repeat the question: After clearing out all of my cabal installed packages, I ran this following session: $ cabal update Downloading the latest package list from hackage.haskell.org james@bast:~/.cabal/packages$ cabal install cabal-dev Resolving dependencies... Downloading cabal-dev-0.9.1... [1 of 1] Compiling Main ( /tmp/cabal-dev-0.9.124882/cabal-dev-0.9.1/Setup.hs, /tmp/cabal-dev-0.9.124882/cabal-dev-0.9.1/dist/setup/Main.o ) Linking /tmp/cabal-dev-0.9.124882/cabal-dev-0.9.1/dist/setup/setup ... Configuring cabal-dev-0.9.1... Warning: This package indirectly depends on multiple versions of the same package. This is highly likely to cause a compile failure. package containers-0.4.2.1 requires array-0.4.0.0 package Cabal-1.14.0 requires array-0.4.0.0 package text-0.11.1.13 requires array-0.4.0.0 package deepseq-1.3.0.0 requires array-0.4.0.0 package containers-0.4.2.1 requires array-0.4.0.0 package HTTP-4000.2.2 requires array-0.4.0.0 package cabal-dev-0.9.1 requires containers-0.4.2.1 package Cabal-1.14.0 requires containers-0.4.2.1 package template-haskell-2.7.0.0 requires containers-0.4.2.1 Building cabal-dev-0.9.1... Preprocessing executable 'ghc-pkg-6_8-compat' for cabal-dev-0.9.1... <command line>: cannot satisfy -package-id Cabal-1.14.0-4af45d3c8d10dc27db38ae0e7e5a952b: Cabal-1.14.0-4af45d3c8d10dc27db38ae0e7e5a952b is unusable due to missing or recursive dependencies: array-0.4.0.0-46f61f5fd9543ebf309552ef84dccc86 containers-0.4.2.1-98f9aa15f9c08b13673dc9d89385f449 (use -v for more information) cabal: Error: some packages failed to install: cabal-dev-0.9.1 failed during the building phase. The exception was: ExitFailure 1 $ So the reason I can't install cabal-dev is apparently either that - it "indirectly depends on multiple versions of the same package." However, cabal does not name the package that it claims cabal-dev requires multiple versions of. - Cabal-1.14.0 has "missing or recursive dependencies", specifically somehow involving array-0.4.0.0 and containers-0.4.2.1. A graph of the dependencies it lists confirms that neither of these claims are true (or the dependencies it lists are false or incomplete): [image: claimed dependency graph of cabal-dev-0.9.1] *So: what am I missing? Who or what is incorrect: me, cabal, or one or more packages?* I am running: $ cabal --version cabal-install version 0.10.2 using version 1.10.1.0 of the Cabal library $ ghc --version The Glorious Glasgow Haskell Compilation System, version 7.4.1 $ Any help with this would be greatly appreciated. I get errors from cabal fairly often and I usually end up giving up because I don't understand the error. James

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Relation between monads and computations
by Dmitriy Matrosov 16 Feb '12

16 Feb '12

Hi, everyone! Not so long ago i started to learn haskell, and now i have a question about relation between monads and computations. In fact, i think, that i understand it and write some explanation for myself, but i'm not sure whether my explanation is correct or not, and will be very thankful if someone check it :) Here it is (i don't know category theory and my math knowledge is poor, so if i accidently use some terms from them incorrectly - it was just my understanding). Monads and computations. Generally computation consists of initial value, computation itself and the result. So it can be illustrated as: type a (M b) b data I ------> C <--------- E f ConstrM I :: a f :: a -> M b data M a = ContrM a | ... E :: b Let's see what happens: i take initial value I of type a and map it to some computation C (of type (M b)) using function f. Then, exist such value E of type b, that C = (ConstrM E). This value E will be the result of computation. Now consider two functions: unitM, which maps value into trivial computation (trivial computation is a computation, which result is equal to initial value): type a (M a) a data I ------> C <--------- I unitM ConstrM I :: a unitM :: a -> M a data M a = ContrM a | ... and bindM, which yields result from one computation, then applies some function f to the result, and makes another computation at the end (E is the result of this last computation). type (M a) a (M b) b data C ---------> I -------> C' <---------- E pattern f ConstrM match data C --------------------> C' <---------- E C `bindM` f ConstrM C :: M a I :: a f :: a -> M b C' :: M b data M a = ContrM a | ... E :: b bindM :: M a -> (a -> M b) -> M b Now, using fucntions unitM and bindM there is possibly to convert arbitrary function (k :: a -> b), which makes from initial value of type a value of type b, into terms of general computation (represented by type M). type a (M a) a b (M b) b data I ------> C ---------> I ------> E -------> C' <-------- E unitM pattern k unitM ConstrM match data I ------> C ---------> I -----------------> C' <-------- E unitM pattern f = (unitM . k) ConstrM match data I ------> C ------------------------------> C' <-------- E unitM `bindM` f ConstrM data I --------------------------------------------> C' <-------- E g = (`bindM` f) . unitM ConstrM I :: a unitM :: a -> M a C :: M a k :: a -> b C' :: M b data M a = ConstrM a | ... E :: b f :: a -> M b bindM :: M a -> (a -> M b) -> M b g :: a -> M b On the first picture i take initial value I of type a, and map it to trivial computation C of type (M a) using unitM. Then i yield result of this trivial computation C (which is I, of course), apply function k to this result (value I) and get value E of type b as result. Then i wrap this value E into trivial computation C' using unitM. Result of computation C' is, of course, E. On the intermediate pictures i show reduction steps of the first picture. And finally i get: i take initial value I of type a, and map it using function g into computation C'. The result of computation C' is value E of type b. So, from arbitrary function (k :: a -> b) i can create function g, which maps initial value I of type a into some computation of type (M b). Actual computation (calculation) will still be performed by k, but the result will have type of general computation (M b) instead of b. Monad is a way to implement such general computation, a way to write a program based on general computations, which later may be redefined, instead of particular ones, which redefinition leads to rewrite of large of amount of code. Monad is a triple of type constructor M and functions unitM and bindM (and errorM, probably?). Type constructor represents computation itself, functions unitM and bindM used to wrap your own specific computation (k :: a -> b) on types a and b into monadic notation (general computation notation). First two of monad laws demand, that unitM maps into trivial computation. ..may be continued.. -- Dmitriy Matrosov

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Effective file I/O with bytestrings
by Johannes Engels 10 Feb '12

10 Feb '12

Dear Haskellers, in several books (RWH, LYAH) I have found the statement that file I/O is more effective with bytestrings than with common strings. As I am doing mostly scientific programming and nearly every program of mine uses file I/O, I would like to check this. So I tried to write a small "benchmark": read a double matrix from file and write it to file again. Regarding to reading, the benefit of bytestrings was actually huge, it was about ten times faster than with strings. What refers to writing, however, I failed completely. Most important, I did not find a function which directly converts doubles to bytestrings. So the best I could figure out was the following ugly workaround using Text.Show.ByteString: import qualified Data.ByteString.Lazy as DL import qualified Text.Show.ByteString as BS import Data.Char import Data.List import Data.Array.Unboxed lineendw8 = DL.pack [fromIntegral (ord '\n')] blankw8 = DL.pack [fromIntegral (ord ' ')] showAll :: UArray (Int, Int) Double -> -- matrix Int -> -- number of rows Int -> -- number of columns DL.ByteString showAll mymatrix numrows numcols = foldr f lineendw8 [0..numrows-1] where f = showLine mymatrix numcols showLine :: UArray (Int,Int) Double -> -- matrix Int -> -- number of columns Int -> -- current row DL.ByteString -> -- accumulator DL.ByteString showLine mymatrix numcols row akku = let f col s = DL.append blankw8 $ DL.append (BS.runPut $ BS.showp (mymatrix!(row,col))) s in DL.append lineendw8 $ foldr f akku [0..numcols-1] main :: IO () main = do -- read file into UArray Int, Int) Double ... -- .... -- ... and write it to file again let bs = showAll mymatrix numrows numcols DL.writeFile "writeOut.dat" bs This was more in order to show goodwill than to present a solution, of course. It actually works, but, ugly as it is, it is by no means faster than the corresponding procedure with strings. So my question: what is the canonical way to write doubles to file? I guess this question must have been posed already hundred times before, so I would also appreciate very much a link to former answers ... Best regards Johannes Engels

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Newtype to avoid orphans
by Adrien Haxaire 06 Feb '12

06 Feb '12

Hello, In the library I am writing, I declare a type Vector for Data.Vector Double. I then create a Num instance for it as it doesn't have one yet. GHC tells me that this instance is an orphan. After reading several answers to issues like mine, I want to get rid of these orphans. The advice I saw mostly is to use the newtype keyword. Is there a way to do it nicely, instead of copying most of the API of Data.Vector ? Now, the only thing I can see is do like in this example: import Data.Vector as V newtype VectorD = VectorD (V.Vector Double) map :: (Double -> Double) -> VectorD -> VectorD map f (VectorD v) = VectorD $ V.map f v Is there a better way ? Best regards, Adrien

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