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Right-associating infix application operators
by Tom Hobbs 06 Jul '10

06 Jul '10

In people's responses to my serialization questions, I've seen them using $. I didn't know what it was so I've looked it up. Can someone please confirm my understanding of what it does, please? According to http://en.wikibooks.org/wiki/Haskell/Practical_monads, after the second code sample in the "Return Values" section, it seems to suggest that $ is only used to avoid using so many brackets. Which seems to make sense, but looking at it's definition in Prelude I really can't see why it's useful. Yitz gave me the code; fmap (runGet $ readNames n) $ L.hGetContents h So can I rewrite this without the $ like this? fmap (runGet (readNames n)) (L.hGetContents h) Is there any additional benefit to using $ than just not having to write as many brackets? Thanks, Tom

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Re: [Haskell-beginners] Right-associating infix application operators
by Antoine Latter 06 Jul '10

06 Jul '10

Hi Tom, There are no additional benifits to the '$' function - you pretty much have it. Wait - one more. It can be used as a 'section' like any other binary operator, so ($ 4) :: (Int -> a) -> a. But that doesn't come up a whole lot. This is much like how (* 5) 6 == 30. Antoine On Jul 6, 2010 6:00 AM, "Tom Hobbs" <tvhobbs(a)googlemail.com> wrote: In people's responses to my serialization questions, I've seen them using $. I didn't know what it was so I've looked it up. Can someone please confirm my understanding of what it does, please? According to http://en.wikibooks.org/wiki/Haskell/Practical_monads, after the second code sample in the "Return Values" section, it seems to suggest that $ is only used to avoid using so many brackets. Which seems to make sense, but looking at it's definition in Prelude I really can't see why it's useful. Yitz gave me the code; fmap (runGet $ readNames n) $ L.hGetContents h So can I rewrite this without the $ like this? fmap (runGet (readNames n)) (L.hGetContents h) Is there any additional benefit to using $ than just not having to write as many brackets? Thanks, Tom _______________________________________________ Beginners mailing list Beginners(a)haskell.org http://www.haskell.org/mailman/listinfo/beginners

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Fwd: [Haskell-beginners] More Deserialization Woes
by Tom Hobbs 06 Jul '10

06 Jul '10

Sorry, I meant to send this to the list, rather than just to Stephen (sorry for the Spam). Tom ---------- Forwarded message ---------- From: Tom Hobbs <tvhobbs(a)googlemail.com> Date: Tue, Jul 6, 2010 at 10:12 AM Subject: Re: [Haskell-beginners] More Deserialization Woes To: Stephen Tetley <stephen.tetley(a)gmail.com> Hello again, I've been reading through various tutorials and they all put IO as the outermost monad, like you suggest. However, I don't think that's what I want. My IO operation is reading from a network stream, not a file, so I want a failure halfway through should signify a complete failure. So, I believe that the type signature for my function should be; ping :: String -> PortNumber -> Maybe (IO [String]) because the result "Just []" is a valid one and does not signify a failure. However, the result of "IO [Just "a", Just "b", Nothing, Nothing]" would signify that communication failed halfway through and would not make sense in my context. This is what the advice seems to be suggesting I write. But in this case, I'd prefer to return "Nothing" to signify that a problem occurred. So my first question is; because I want to do something so different from the majority of the articles I've read; am I in a niche where my requirement makes sense, or does my requirement make no sense - a theory that is backed up by the fact that no one else seems to be doing that... Now, I'm not sure I can get there by myself, since I'm struggling to get the right incantation of catching errors but I'll keep plugging away at that for a while. But can someone help me with my next question also. Is it possible, to extract the values out of the IO monad so it can be used in pure functions. For example, once ping has returned it's Maybe [IO String], I would like to be able to create another function such as; purePing :: String -> PortNumber -> Maybe [String] purePing a p = removeIOMonad (ping a p) where removeIOMonad Nothing = Nothing removeIOMonad [] = Just [] removeIOMonad (x:xs) = clevelDropIOMagic x : removeIOMonad xs ...or something... Once the IO [String] has been read from the stream, no further IO is necessary, so any other function should be able to just use the list and not worry about the IO stuff. Again my questions are, Is this an okay thing to do, or is my design/idea very, very wrong? Thanks again for the help, Tom On Thu, Jul 1, 2010 at 1:53 PM, Stephen Tetley <stephen.tetley(a)gmail.com>wrote: > Hi Tom > > This is bit that is wrong: > > ((UTF.toString name) : readNames (n-1) h) > > > You are trying to cons (:) a String onto an IO [String]: > > (UTF.toString name) :: String > > (readNames (n-1) h) :: IO [String] > > (:) :: a -> [a] -> [a] > > This is easy mistake to make, I probably still make it myself now and > again if I'm typing faster than I'm thinking. > > As for ping, you are right that the common answer type will be "IO > (Maybe [String])" i.e. > > ping :: String -> PortNumber -> IO (Maybe [String]) > > > Generally you will want IO as the outermost (type-) constructor for an > expression involving IO, this is because you can't 'escape' IO. There > are some exceptions where IO isn't the outermost type constructor, the > common one I can think of is you might want to build a list of IO > actions [IO ()]. You would then commonly pass on this list to evaluate > it later with mapM_. > > > Neil Mitchell's "IO without (concentrating) on monads tutorial" is a > very good place to start: > http://neilmitchell.blogspot.com/2010/01/haskell-io-without-monads.html > > Best wishes > > Stephen >

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Right-associating infix application operators
by Tom Hobbs 06 Jul '10

06 Jul '10

In people's responses to my serialization questions, I've seen them using $. I didn't know what it was so I've looked it up. Can someone please confirm my understanding of what it does, please? According to http://en.wikibooks.org/wiki/Haskell/Practical_monads, after the second code sample in the "Return Values" section, it seems to suggest that $ is only used to avoid using so many brackets. Which seems to make sense, but looking at it's definition in Prelude I really can't see why it's useful. Yitz gave me the code; fmap (runGet $ readNames n) $ L.hGetContents h So can I rewrite this without the $ like this? fmap (runGet (readNames n)) (L.hGetContents h) Is there any additional benefit to using $ than just not having to write as many brackets? Thanks, Tom

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Re: [Haskell-beginners] upgrade Hackage show to QuickCheck 2 for lambdabot
by Antoine Latter 06 Jul '10

06 Jul '10

Including the café. On Jul 2, 2010 8:49 AM, "Mark Wright" <markwright(a)internode.on.net> wrote: Hi, I'm trying to upgrade Hackage show to QuickCheck 2, after applying the diffs below (which may not be correct, since I am a beginner), I am left which this error message: runghc ./Setup.hs build Preprocessing library show-0.3.4... Building show-0.3.4... [4 of 4] Compiling ShowQ ( ShowQ.hs, dist/build/ShowQ.o ) ShowQ.hs:104:20: Not in scope: `generate' Compilation exited abnormally with code 1 at Fri Jul 2 23:07:17 The error occurs in this method: tests :: Gen Result -> StdGen -> Int -> Int -> [[String]] -> IO String tests gen rnd0 ntest nfail stamps | ntest == 500 = done "OK, passed" ntest stamps | nfail == 1000 = done "Arguments exhausted after" ntest stamps | otherwise = case ok result of Nothing -> tests gen rnd1 ntest (nfail+1) stamps Just True -> tests gen rnd1 (ntest+1) nfail (stamp result:stamps) Just False -> return $ "Falsifiable, after " ++ show ntest ++ " tests:\n" ++ reason result where result = generate (((+ 3) . (`div` 2)) ntest) rnd2 gen (rnd1,rnd2) = split rnd0 The QuickCheck 1 generate method is near the bottom this page: http://hackage.haskell.org/packages/archive/QuickCheck/1.2.0.0/doc/html/Tes… but I can not find generate in QuickCheck 2. I am wondering if you have any ideas on how to fix it? I'm trying to package lambdabot on Solaris. I have already packaged the Haskell Platform and about 90 packages, they are in: http://pkgbuild.sourceforge.net/spec-files-extra/ Thanks very much, Mark here are the diffs: goanna% diff -wc show-0.3.4-orig/ShowQ.hs show-0.3.4/ShowQ.hs *** show-0.3.4-orig/ShowQ.hs Wed Jan 20 11:24:11 2010 --- show-0.3.4/ShowQ.hs Fri Jul 2 23:07:13 2010 *************** *** 12,22 **** --- 12,25 ---- import qualified Test.SmallCheck (smallCheck, Testable) import Test.QuickCheck + import Test.QuickCheck.Arbitrary import Data.Char import Data.List import Data.Word import Data.Int import System.Random + import Control.Exception (evaluate) + import Test.QuickCheck.Property (ok, stamp) type T = [Int] type I = Int *************** *** 23,36 **** --- 26,45 ---- instance Arbitrary Char where arbitrary = choose (minBound, maxBound) + + instance CoArbitrary Char where coarbitrary c = variant (ord c `rem` 4) instance Arbitrary Word8 where arbitrary = choose (minBound, maxBound) + + instance CoArbitrary Word8 where coarbitrary c = variant (fromIntegral ((fromIntegral c) `rem` 4)) instance Arbitrary Ordering where arbitrary = elements [LT,EQ,GT] + + instance CoArbitrary Ordering where coarbitrary LT = variant 1 coarbitrary EQ = variant 2 coarbitrary GT = variant 0 *************** *** 37,42 **** --- 46,53 ---- instance Arbitrary Int64 where arbitrary = sized $ \n -> choose (-fromIntegral n,fromIntegral n) + + instance CoArbitrary Int64 where coarbitrary n = variant (fromIntegral (if n >= 0 then 2*n else 2*(-n) + 1)) instance (Integral a, Arbitrary a) => Arbitrary (Ratio a) where *************** *** 48,53 **** --- 59,65 ---- else (b % a) else (a % b) + instance (Integral a, CoArbitrary a) => CoArbitrary (Ratio a) where coarbitrary m = variant (fromIntegral $ if n >= 0 then 2*n else 2*(-n) + 1) where n = numerator m *************** *** 87,93 **** Just False -> return $ "Falsifiable, after " ++ show ntest ++ " tests:\n" ! ++ unlines (arguments result) where result = generate (((+ 3) . (`div` 2)) ntest) rnd2 gen (rnd1,rnd2) = split rnd0 --- 99,105 ---- Just False -> return $ "Falsifiable, after " ++ show ntest ++ " tests:\n" ! ++ reason result where result = generate (((+ 3) . (`div` 2)) ntest) rnd2 gen (rnd1,rnd2) = split rnd0 goanna% _______________________________________________ Beginners mailing list Beginners(a)haskell.org http://www.haskell.org/mailman/listinfo/beginners

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problem with System.Directory.Tree
by Anand Mitra 06 Jul '10

06 Jul '10

Hello All, I want to build a program which will recursively scan a directory and build md5sum for all the files. The intent is to do something similar to unison but more specific to my requirements. I am having trouble in the initial part of building the md5sums. I did some digging around and found that "System.Directory.Tree" is a very close match for what I want to do. In fact after a little poking around I could do exactly what I wanted. ,---- | import Monad | import System.Directory.Tree | import System.Directory | import Data.Digest.Pure.MD5 | import qualified Data.ByteString.Lazy.Char8 as L | | calcMD5 = | readDirectoryWith (\x-> liftM md5 (L.readFile x)) `---- This work perfectly for small directories. readDirectoryWith is already defined in the library and exactly what we want ,---- | *Main> calcMD5 "/home/mitra/Desktop/" | | "/home/mitra" :/ Dir {name = "Desktop", contents = [File {name = | "060_LocalMirror_Workflow.t.10.2.62.9.log", file = | f687ad04bc64674134e55c9d2a06902a},File {name = "cmd_run", file = | 6f334f302b5c0d2028adeff81bf2a0d9},File {name = "cmd_run~", `---- However when ever I give it something more challenging it gets into trouble. ,---- | *Main> calcMD5 "/home/mitra/laptop/" | *** Exception: /home/mitra/laptop/ell/calc-2.02f/calc.info-27: | openFile: resource exhausted (Too many open files) | *Main> 29~ `---- If I understand what is happening it seems to be doing all the opens before consuming them via md5. This works fine for small directories but for any practical setup this could potentially be very large. I tried forcing the md5 evaluation in the hope that the file descriptor will be freed once the entire file is read. That did not help, either because I could not get it right or there is some more subtle I am missing. I also had a look at the code in module "System.Directory.Tree" and although it gave me some understanding of how it works I am no closer to a solution. regards -- Anand Mitra

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Re: [Haskell-beginners] 'cabal install hdirect' can't find hdirect
by Larry Evans 05 Jul '10

05 Jul '10

On 07/04/10 07:28, Daniel Fischer wrote: > On Sunday 04 July 2010 13:45:01 you wrote: > >> src/Lex.lhs:330:8: >> Illegal signature in pattern: Int >> Use -XPatternSignatures to permit it >> cabal: Error: some packages failed to install: >> hdirect-0.21.0 failed during the building phase. The exception was: >> exit: ExitFailure 1 >> ~/download/haskell/libs $ which happy >> /home/evansl/.cabal/bin/happy >> ~/download/haskell/libs $ >> >> Any ideas about how to solve this problem? >> > > Sure. First, tell the maintainer, he might want to fix it for all. > Done. > Then, the quick fix for you, > * unpack the .tar.gz > * change the version in the .cabal file (my preferred method is appending a > ".1" to the version number). You need that in case you want to build some > library which depends on hdirect, so that cabal-install picks your bumped > newer version rather than the newest version on hackage. > * make it compile, e.g. by adding a field > Extensions: PatternSignatures > to the cabal file, or by putting a > {-# LANGUAGE PatternSignatures #-} > pragma to the offending file. > Then run cabal install in the unpacked and modified directory. > Several iterations of edit file; cabal install may be necessary. > > Downloaded: http://hackage.haskell.org/packages/archive/hdirect/0.21.0/hdirect-0.21.0.t… then unzipped and untarred it then edited hdirect.cabal by adding: Extensions: PatternSignatures Below the line: library However, still got: src/Desugar.lhs:2053:1: Warning: Pattern match(es) are non-exhaustive In a case alternative: Patterns not matched: (_, []) (_, IDL.IncludeEnd : _) (_, (IDL.IncludeStart _) : _) (_, (IDL.Pragma _) : _) ... [60 of 65] Compiling IDLToken ( src/IDLToken.lhs, dist/build/hdirect/hdirect-tmp/IDLToken.o ) [61 of 65] Compiling LexM ( src/LexM.lhs, dist/build/hdirect/hdirect-tmp/LexM.o ) [62 of 65] Compiling Lex ( src/Lex.lhs, dist/build/hdirect/hdirect-tmp/Lex.o ) src/Lex.lhs:330:8: Illegal signature in pattern: Int Use -XPatternSignatures to permit it cabal: Error: some packages failed to install: Any further help is appreciated. -regards, Larry

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threadDelay does not delay thread execution?
by Benjamin Edwards 05 Jul '10

05 Jul '10

Hi All, I am looking to make a simple program that polls an interface once a second or so. I thought I could use threadDelay for this, so knocked up the following toy program to test my idea. module TD where import Control.Concurrent import Control.Concurrent.MVar main :: IO () main = do mv <- newEmptyMVar forkIO $ loop mv 0 "Hai" takeMVar mv return () where loop mv n msg = do putStrLn msg threadDelay 10000 if n < 10 then loop mv (n+1) msg else putMVar mv () It doesn't work, in as much as I see all the messages printed seemingly as fast as Std out will print them. What am I doing wrong? All help is much appreciated, Ben

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curry in a hurry
by prad 04 Jul '10

04 Jul '10

today i'm trying to understand curry and uncurry (in 3 parts with specific questions labelled as subsections of the parts). part 1 if we have f (x,y) = x + y this is a function working on a single argument (x,y) albeit composed of 2 parameters. however, since all functions are curried in haskell (read that here: http://www.haskell.org/haskellwiki/Currying) what is really happening is (f x) y or specific to this case (+ x) y since f ends up being (+) as defined. now if we write g = curry f we are naming a function whose purpose is to (f x) whatever comes its way allowing us to do neat things like: map (g 3) [4,5,6] [7,8,9] which we can't do by map (f 3) [4,5,6] because f :: (Num t) => (t, t) -> t meaning f - is a function ... the (Num t) - applies itself ... the => - to a Pair of (Num t)s - gives back a Num t 1.1 am i reading the type statement correctly? while g :: Integer -> Integer -> Integer means g takes an Integer, applies itself and that Integer to another Integer and computes another Integer. 1.2 how come these are Integer suddenly and not Num t? this is a problem because while i can do f (2.3,9.3) i get an error when i try (g 2.3) 9.3 ghci wants an instance declaration for (Fractional Integer) which puzzles me because g came about through currying f which is fine with fractions. part 2 now the next discovery is really strange to me. if i name f x y = x + y we see f :: (Num a) => a -> a -> a which looks like the curried form of f (x,y) in fact that's what it exactly is and i can do map (f 3) [4,5,6] [7,8,9] just as i did with g before!! which is what the wiki statement says too: f :: a -> b -> c is the curried form of g :: (a, b) -> c however, it starts by stating: "Currying is the process of transforming a function that takes multiple arguments into a function that takes just a single argument and returns another function if any arguments are still needed." http://www.haskell.org/haskellwiki/Currying 2.1 so does all this mean that f (x,y) is the function that takes multiple arguments and not a single argument as i initially thought and f x y is the function that actually takes a single argument twice? part 3 some of the above seems to be confirmed by looking at these types c x y = x + y c :: (Num a) => a -> a -> a so that's curried u (x,y) = x + y u :: (Num t) => (t,t) -> t so that's uncurried :t uncurry c uncurry c :: (Num a) => (a, a) -> a :t curry u curry u :: (Num a) => a -> a -> a but :t uncurry u uncurry u :: (Num (b -> c)) => ((b -> c, b -> c), b) -> c we're trying to uncurry something that is already uncurried and :t curry c curry c :: (Num (a, b)) => a -> b -> (a, b) -> (a, b) we're trying to curry something that is already curried 3.1 just what are these strange looking things and how should their types be interpreted? -- In friendship, prad ... with you on your journey Towards Freedom http://www.towardsfreedom.com (website) Information, Inspiration, Imagination - truly a site for soaring I's

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'cabal install hdirect' can't find hdirect
by Larry Evans 04 Jul '10

04 Jul '10

hdirect is here: http://hackage.haskell.org/package/hdirect however, as shown by the following terminal session, cabal can't find it. What should I do to insteall hdirect? ~/download/haskell/libs $ cabal -V cabal-install version 0.6.0 using version 1.6.0.1 of the Cabal library ~/download/haskell/libs $ uname -a Linux evansl-desktop 2.6.24-24-generic #1 SMP Fri Sep 18 16:16:18 UTC 2009 x86_64 GNU/Linux ~/download/haskell/libs $ cabal install happy --dry-run Resolving dependencies... In order, the following would be installed (use -v for more details): mtl-1.1.0.2 happy-1.18.1 ~/download/haskell/libs $ cabal install hdirect --dry-run cabal: There is no package named hdirect ~/download/haskell/libs $

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