January 2011 - Beginners

Trouble with liftIO in ReaderT
by John Smith 30 Jan '11

30 Jan '11

I'm trying to attach a handler to a Gtk2Hs event, and get the error: No instance for (Control.Monad.IO.Class.MonadIO (mtl-1.1.0.2:Control.Monad.Reader.ReaderT (GHC.Ptr.Ptr EAny) IO)) arising from a use of `liftIO' Possible fix: add an instance declaration for (Control.Monad.IO.Class.MonadIO (mtl-1.1.0.2:Control.Monad.Reader.ReaderT (GHC.Ptr.Ptr EAny) IO)) This occurs with the sample at http://www.haskell.org/haskellwiki/Gtk2Hs/Tutorials/Intro import Graphics.UI.Gtk import Control.Monad.Trans(liftIO) main = do initGUI window <- windowNew window `on` deleteEvent $ liftIO mainQuit >> return False -- i.e., on window deleteEvent (liftIO mainQuit >> return False) widgetShow window mainGUI Why can't GHC find the instance MonadIO m => MonadIO (ReaderT r m)?

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[Haskell-Beginners] Laziness & referential transparency
by Tim Baumgartner 29 Jan '11

29 Jan '11

Hi everybody, after reading a post by Don Stewart on http://cgi.cse.unsw.edu.au/~dons/blog/2008/05/16#fast, I was not really satisfied with the necessity to rewrite mean xs = sum xs / fromIntegral (length xs) to an explicitly recursive function. I investigated a bit further and got really confused after the following GHCi session: Prelude> length [1..10^8] + length [1..10^8] 200000000 Prelude> let xs = [1..10^8] Prelude> length xs + length xs <interactive>: out of memory Can someone explain this and give a simple, reasonable rule by which I can predict such things in the future? Isn't it a pitfall? Tim

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A very counterintuitive behaviour of Haskell
by Renzo Orsini 28 Jan '11

28 Jan '11

In studying Haskell, I produced the following output from GHC: xxx-3:~ xxx$ GHCi GHCi, version 6.12.3: http://www.haskell.org/ghc/ :? for help Loading package ghc-prim ... linking ... done. Loading package integer-gmp ... linking ... done. Loading package base ... linking ... done. Loading package ffi-1.0 ... linking ... done. Prelude> let f 7 = "ok" Prelude> let f x = "no" Prelude> f 3 "no" Prelude> f 7 "no" I suppose it is correct. However, for someone who is interested in the language, it seems very counterintuitive... Somebody would be so kind to explain to a neophyte this "feature" of the language? Thank you very much. Renzo

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Re: [Haskell-beginners] A very counterintuitive behaviour of Haskell
by Renzo Orsini 27 Jan '11

27 Jan '11

Thank you very much to everybody. Your explanation is perfectly clear! I was reading the tutorial at http://learnyouahaskell.com/syntax-in-functions . Initially the tutorial cited GHCi, so I mistakenly wrote the functions on different lines. Thank very much again. I regained the desire of continuing my study... :-) Renzo On Jan 27, 2011, at 16:09 , jean verdier wrote: > You have defined 2 functions that are called f and only the last > definition is used (f x = "no"). > The definition you want should be written > let f 7 = "ok"; f x = "no" > so the function is defined once. > The problem comes from using the interpreter and not from haskell. > > > On Thu, 2011-01-27 at 15:55 +0100, Renzo Orsini wrote: >> In studying Haskell, I produced the following output from GHC: >> >> xxx-3:~ xxx$ GHCi >> GHCi, version 6.12.3: http://www.haskell.org/ghc/ :? for help >> Loading package ghc-prim ... linking ... done. >> Loading package integer-gmp ... linking ... done. >> Loading package base ... linking ... done. >> Loading package ffi-1.0 ... linking ... done. >> Prelude> let f 7 = "ok" >> Prelude> let f x = "no" >> Prelude> f 3 >> "no" >> Prelude> f 7 >> "no" >> >> >> I suppose it is correct. However, for someone who is interested in the language, it seems very counterintuitive... Somebody would be so kind to explain to a neophyte this "feature" of the language? >> >> Thank you very much. >> >> Renzo >> _______________________________________________ >> Beginners mailing list >> Beginners(a)haskell.org >> http://www.haskell.org/mailman/listinfo/beginners > >

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Trying to use map
by Michael Litchard 27 Jan '11

27 Jan '11

Here's what I'm working with, followed by what I am trying to do, and the type error I get. I'm leaving things out that I do not think are important. Let me know if I'm missing nessecary info. > curlResp > :: (MonadError String m, MonadIO m) > => Curl -> URLString -> [CurlOption] -> m String --CurlResponse > curlResp curl url opts = do > resp <- liftIO $ (do_curl_ curl url opts :: IO CurlResponse) > let code = respCurlCode resp > status = respStatus resp > if code /= CurlOK || status /= 200 > then throwError $ "Error: " ++ show code ++ " -- " ++ show status > else return $ respBody resp > screenScraping :: String -> [String] > screenScraping responseBody = > let collectedStrings = processHTML responseBody > collectedIDLists = createIDList collectedStrings > in constructedResourceURIs urlBase collectedIDLists > resourceOpts :: [CurlOption] > resourceOpts = > [ CurlHttpHeaders > [ "Accept: text/javascript, text/html, application/xml, text/xml, */*" > , "Accept-Language: en-us,en;q=0.5" > , "Accept-Charset: ISO-8859-1,utf-8;q=0.7,*;q=0.7" > , "Keep-Alive: 115" > , "Connection: keep-alive" > , "X-Requested-With: XMLHttpRequest" > , "X-Prototype-Version: 1.6.0.3" > ] > , CurlEncoding "gzip,deflate" > ] > obtainCookies :: IO Curl -> String -> IO () > obtainCookies curl responseBody = do > curl' <- curl > let collectedResources = screenScraping responseBody > in mapM ( curlResp curl' resourceOpts) collectedResources > return () > main :: IO () > main = do > curl <- initCurl > user:pass:_ <- getArgs > resp <- generateResourceHtml user pass > > case resp of > Left err -> print err > Right body -> obtainCookies curl body Here's the error I get. Couldn't match expected type `Char' against inferred type `CurlOption' Expected type: URLString Inferred type: [CurlOption] In the second argument of `curlResp', namely `resourceOpts' In the first argument of `mapM', namely `(curlResp curl' resourceOpts)' The problem I see is a misformed mapM. I am trying to do something like this curlResp curl' resourceOpts "-here-be-a-url" where collectedResources is a [String]. Not sure how to map over it correctly. The other problem I see is this > obtainCookies :: IO Curl -> String -> IO () > obtainCookies curl responseBody = do > curl' <- curl > let collectedResources = screenScraping responseBody > in mapM ( curlResp curl' resourceOpts) collectedResources > return () this function does a monadic action (all I want is the cookies) and I don't need the return value. I am not sure that the final line return (), is what I want. My primary question is this. how do I map over collectedResources correctly? Secondary question, is the return correct?

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Re: [Haskell-beginners] Looking for some guidance to installing GHCI on MAC
by patricklynch 26 Jan '11

26 Jan '11

...i tried the March 2010 installation on the MAC and I get the same result...if there is a requirement for Xcode, shouldn't it be listed in the Installation Instructions... I tried another installation but it too required Xcode... ...again, I was able to install GHC on a PC and was able to install HUGs on the Mac... ...really frustrating... I'm not going to pay Apple $99, so I may have to give up on the Mac for a while - bummer... -----Original Message----- From: patricklynch [mailto:kmandpjlynch@verizon.net] Sent: Sunday, January 23, 2011 4:44 PM To: 'Antoine Latter' Cc: 'Wes Morgan'; 'beginners(a)haskell.org' Subject: RE: [Haskell-beginners] Looking for some guidance to installing GHCI on MAC ...sorry for being a 'klutz' - but I'm new to the mac too... ...i'm guessing that I have Xcode installed...i see an X icon that has an X11 label - when I click it, it displays nothing...bummer... ...since I was able to install Hugs I'm guessing that Xcode is installed...however, I have no way of knowing if this is true...and Apple wants $99 for a Developer fee, bummer...I was not given an installation disk... I went to the site: haskage.haskell.org\platform\mac.html... It gives me a three step installation procedure... ...the 1st deletes any previous installation of GHC - this worked... ...the 2nd gets me to a screen that states Standard install on "Macintosh HD"; with a message: click Install...{however, the Install Button can not be clicked}...this is as far as I can get... -----Original Message----- From: Antoine Latter [mailto:aslatter@gmail.com] Sent: Sunday, January 23, 2011 2:55 PM To: patricklynch Cc: Wes Morgan; beginners(a)haskell.org Subject: Re: [Haskell-beginners] Looking for some guidance to installing GHCI on MAC The problem I had at first with the Haskell Platform installer was that I didn't have Xcode installed. This should be on one of the OS X install discs, I think. You can also download it from Apple: http://developer.apple.com/technologies/xcode.html It might be easier if you could send us the error message you're getting. Antoine On Sun, Jan 23, 2011 at 2:18 PM, patricklynch <kmandpjlynch(a)verizon.net> wrote: > ...i tried to use the installation for Mac at Haskell.org, but it gives me > an error message [something like: call your software vendor]... > > ...i tried the Homebrew, see email below, but it wants my Apple Developer > id...i don't have one... > > ...i installed it on my pc and all it took was a single keystroke... > ...i installed Hugs on my Mac without any problem.. > > ...i'd appreciate any help, I really need to work with ghc on my mac > > Go Jets > > -----Original Message----- > From: beginners-bounces(a)haskell.org [mailto:beginners-bounces@haskell.org] > On Behalf Of Wes Morgan > Sent: Sunday, January 23, 2011 7:36 AM > To: beginners(a)haskell.org > Subject: Re: [Haskell-beginners] Looking for some guidance to installing > GHCI on MAC > > I just used Homebrew (http://mxcl.github.com/homebrew/) Once that's > installed, 'brew install ghc' gets you GHC, GHCi, cabal, etc. > > Wes > > On Jan 23, 2011, at 6:00 AM, "beginners-request(a)haskell.org" > <beginners-request(a)haskell.org> wrote: > >> Re: [Haskell-beginners] Looking for some guidance to >> installing GHCI on MAC > > _______________________________________________ > Beginners mailing list > Beginners(a)haskell.org > http://www.haskell.org/mailman/listinfo/beginners > > > _______________________________________________ > Beginners mailing list > Beginners(a)haskell.org > http://www.haskell.org/mailman/listinfo/beginners >

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Bytestring question
by Peter Braun 26 Jan '11

26 Jan '11

Hi everyone, as an exercise for learning Haskell i'm writing a program that converts Ascii Stl files (a simple format for 3D model data) into binary Stl format. In my first attempt i used normal strings and the result was therefore very slow. Now i rewrote the program to use lazy bytestrings instead. But well... it got even slower, so i'm probably doing something terribly wrong ;) Here's what i do (the relevant parts): ... asciiFile <- L.readFile (args!!0) binHandle <- openBinaryFile (args!!1) WriteMode let asciiLines = L.split (c2w '\n') asciiFile ... parseFile binHandle (Normal, tail asciiLines) -- First line contains a comment ... where L is Data.ByteString.Lazy. readFile ought to be lazy so it should not read the whole file into ram at this point. But when i split the lines and pass them to a function, is this still carried out lazily? parseFile processes a line, depending on the StlLineType and then calls itself recursive like this: parseFile :: Handle -> (StlLineType, [L.ByteString]) -> IO () ... parseFile h (Vertex1, s) = do let vals = extractVertex (head s) L.hPutStr h $ runPut (writeFloatArray vals) parseFile h (Vertex2, tail s) ... extractVertex looks like this: extractVertex :: L.ByteString -> [Float] extractVertex s = let fracs = filter (\n -> L.length n > 0) $ L.split (c2w ' ') s in [read (C.unpack(fracs!!1)) :: Float, read (C.unpack(fracs!!2)) :: Float, read (C.unpack(fracs!!3)) :: Float] where C is Data.ByteString.Lazy.Char8. It splits a byte string, filters out the whitespaces and converts certain entries to floats. Maybe unpack is an expensive operation. Is there a better way to convert a Bytestring to float? I know, this is bad Haskell code ;) But where is my grand, obvious misuse of Bytestring? I'm grateful for any suggestion to improve that code. I'm using ghc, version 6.12.1. Thank you, Peter

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Wildcards in expressions
by Tom Murphy 26 Jan '11

26 Jan '11

Hi everyone, Any suggestions? Given that someList :: [(a,b,String)] , the expression (_, _, "Tom") `elem` someList seems so intuitive that, even though it doesn't work, I've been searching for several hours for something similar. Do any of you know a way I can pull this off? I know there are plenty of ways to get the functionality, but something similarly intuitive and succinct would be great. Thanks for the help, Tom

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Please help with some HXT code
by Jeff Lasslett 26 Jan '11

26 Jan '11

Hello, I am attempting to use HXT to scraped some html from the web. In my html are a few <div>s, all with a particular attribute. I wish the get the text nodes out of these <div>s. For each <div> I can use listA ( deep isText >>> getText ) >>> arr concat to get all the text children an dmake one string out of them. I'm just struggling to get all the text out of many similar <div>s. Here's an html fragment similar to what I'm trying to process: <tr> <td> <div attr="value"> the quick <br/> brown fox </div> </td> <td> <div attr="value"> jumps over<br/> </div> </td> </tr> <tr> <td> <div attr="value"> the lazy dog </div> </td> </tr> >From that lot I would like the string "the quick brown fox jumps over the lazy dog". How might this be achieved??? Thanks, Jeff

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Function result being inversed
by Xavier Shay 25 Jan '11

25 Jan '11

Hello, I am confused by the following code. I would expect results of True, False. $ ghci *Main> let f x = x 4 *Main> f(<) 3 False *Main> f(<) 5 True This came about because I was trying to refactor a sort function I wrote: mySort [] = [] mySort (h:t) = (f (<= h)) ++ [h] ++ (f (> h)) where f x = mySort (filter x t) I came up with this, which appears to work, though the comparison operators are backwards. mySort [] = [] mySort (h:t) = f(>) ++ [h] ++ f(<=) where f x = mySort (filter (x h) t) Cheers, Xavier

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