- Beginners - Haskell.org

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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scanl and infinite sequences [scanl for Fibonacci sequences]
by Seung Soo, Ha 25 Jan '11

25 Jan '11

I was playing with the project Euler problems and found heaps of Fibonacci generating codes in Haskell. One caught my eye: fib = 1 : scanl (+) 1 fib ....(1) when my understanding was limited to "folds just replace : with the function you provide scans show all the intermediate states of your fold" this line of code was very easy to understand. Efficient too(according to the wiki article on Haskell) But now that I'm trying to grasp foldl and foldr, I no longer understand that line of code!!! This is especially confusing, as afaik, (1) is generating an infinite list but I thought you can't use foldl for infinite lists! Tried replacing scanl with scanr (foldr is supposed to work with infinite lists right?) fib = 1 : scanr (+) 1 fib .....(2) somewhat surprisingly, (2) crashes. I've read numerous articles on folds on the haskell wiki, and also this stack overflow post (http://stackoverflow.com/questions/3082324) Just as I thought I was getting a hold on the concept of folds... This.. What am I missing here?

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Bug or Intended behavior? ( Exception: stack overflow )
by Seung Soo, Ha 24 Jan '11

24 Jan '11

my setup: phenom II quadcore, 4GB mem Ubuntu Maverick While implementing my own reverse in "99 haskell questions", I made the following mistake: (Caution: stack over flow causing code below) reverse'::[a] -> [a] reverse' [] = [] reverse' [x] = [x] reverse' xs = last xs : init (reverse' xs) The last line is the wrong part. It should be reverse' xs = last xs : reverse' (init xs) anyway, the original code seems to cause an infinite recursion by repeating the last part forever. When I try this out on ghci: *Main> reverse' [1,2,3,4,5] [5*** Exception: stack overflow *Main> The problem is, unless I manually exit ghci, the "stack" does not seem to be freed (garbage collected?) and memory remains fully occupied by ghci. I tried waiting a bit, but it stayed the same. Is this expected behavior? Shouldn't the GC kick in? Is this orthogonal of the GC? I think this would help me understand haskell and GHC much better.

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