- Beginners - Haskell.org

random
by John Moore 24 Oct '09

24 Oct '09

Hi All, Can anyone help me I want to produce a list of three random numbers for e.g. [7,8,1] I tried using x <- getStdRandom $ randomR (1,10) but don't really understand this and it only generates one number. Any help greatly appreciated. Regards John

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using quickcheck for blackbox testing for 3rd party apps.
by Srikanth K 24 Oct '09

24 Oct '09

Hi All, I am exploring use of quickCheck to define the test-specification for a 3rd party component. The component is basically an Command/Response type of application. Hence to validate it, I thought that the best way might be to take the following approach(based on the quickCheckM paper) a. describe the component in a purely functional way. The functions would take the "AppState", "Command" as parameters and would return a Response. Let me call this the "Abstract Application" The signature for this would therefore be of the form abstract_App :: AppState -> Command -> Response b. define an agent (Let me call it as Real_App) that that would (i) take a initial State, "Command" and send it to the "Application Under test" using some IPC mechanisms(probably socket or FIFO files) (ii) parse the received response and send it back as "Response". As a result the the API for the agent would be of the form real_APP :: AppState -> Command -> IO Response c. write a quick-check specification that would send a bunch of commands to the "Abstract Application" and check for equivalence with the value returned by the "real_App" However, being a newbie, I am not able to understand how I can extract the "Response" part from the IO Response. I want something as follows prop_App state_x command_y = (abstract_App state_x command_y ) == (real_App state_x command_y) The issue here is that since the reall_App would return a "IO Response" and not "Response" I cannot really compare the abstract_App with the real_App. Any thought on how I can extract the value from the "IO Response" so that I can compare the the two implementations for equvivalence. FYI, I have gone through the various quickcheck related docs like http://www.cs.chalmers.se/~rjmh/QuickCheck/ http://www.haskell.org/haskellwiki/QuickCheck_as_a_test_set_generator Thanks - Srikanth

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Caching evaluation of lazy lists
by Philip Scott 23 Oct '09

23 Oct '09

Hi folks, Quick question for you. Suppose I have some lazy list, x, which is very expensive to evaluate, and I want to do something with it more than once; for example: f x = g (take 5 x) (take 6 x) Is Haskell clever enough to only evaluate the list as much as is needed, or will it evaluate it once to get the first five values, and again to get the first 6 (when really it only needed to get one more). Or is it really really clever, and works out it needs to take 6 to begin with, then just give you the first 5 for the first call? More tricky - is there a way to make that cache (if it exsists) persist in an interactive session? Eventually I am intending to write my own application with a little console at the bottom which does some ghci-esque magic, but for now lets say I am in ghci and I want to call f again with the same list. Is there a way to avoid it from forcing a recomputation of my lazy list? Any advice greatly appreciated, Philip

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\x -> x < 0.5 && x > -0.5
by Michael Mossey 23 Oct '09

23 Oct '09

Is there a nifty way to write filter (\x -> x < 0.5 && x > -0.5) xs without explicitly using x? Maybe arrows? I have a vague understanding that arrows can "send" an argument to more than one computation. -Mike

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Haskell Output Help
by Chandni Navani 23 Oct '09

23 Oct '09

I have a list of lists which all contain strings. [[String]]. I need to figure out how to print them so that after each individual string, there is a new line. If this is the initial list [["abc", "cde"] ["fgh", "ghi"]] [["abc" "cde"] ["fgh", "ghi"]] Can anyone help me figure this out? Thanks.

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Stack overflow, but hard to understand
by Michael Mossey 21 Oct '09

21 Oct '09

Okay, beginner encountering first major bug. I'm getting a stack overflow on a program that uses a lot of laziness to construct and modify lists---but the overflow happens even if I take just one element from the final list! The overflow seems to have started when I added the functions 'gauss' and 'gaussList' below, but I am at a loss to know why---these functions work perfectly in test cases and can even generate a long list. The following program is distilled (with a lot of work!) down to the essence of what triggers the problem. import Control.Monad import Control.Monad.Random import Data.List -- Here I put a moderate-sized list of filenames which just generic -- text or code. The longer the list, -- the more likely to overflow stack. Which is curious, because if IO were -- lazy here, it shouldn't be reading more than the first character of the -- first file. Methinks a hint to the problem. fileList = ... data TextGroup = Single Char Float deriving (Show) textGroup_addDelay :: TextGroup -> Float -> TextGroup textGroup_addDelay (Single c del) del2 = Single c (del+del2) mkTextGroup :: Char -> Rand StdGen TextGroup mkTextGroup c = liftM (Single c) gauss -- Here's the kinda weird thing I'm doing with Rand. I want to -- use fromList as the first computation to randomly choose -- a second Rand computation. It turns out if you replace -- this funkiness with -- gauss = getRandomR (0,1) -- the whole thing works. gaussList :: Rand StdGen (Rand StdGen Float) gaussList = fromList [ (getRandomR( -1.0 ,-0.8 ), 2) , (getRandomR( -0.8 ,-0.6 ), 2) , (getRandomR( -0.6 ,-0.4 ), 4) , (getRandomR( -0.4 ,-0.2 ), 8) , (getRandomR( -0.2 , 0.0 ), 12) , (getRandomR( 0.0 , 0.2 ), 12) , (getRandomR( 0.2 , 0.4 ), 8) , (getRandomR( 0.4 , 0.6 ), 4) , (getRandomR( 0.6 , 0.8 ), 2) , (getRandomR( 0.8 , 1.0 ), 2) ] gauss :: Rand StdGen Float gauss = do m <- gaussList m main = do gen0 <- newStdGen bufs <- mapM readFile fileList let buf = concat bufs let tgs = evalRand (do orig <- mapM mkTextGroup buf addBreaks orig) gen0 writeFile "output.ank" (show $ take 1 tgs)

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haskell on snow leopard
by Bernardas Jankauskas 20 Oct '09

20 Oct '09

Hello, For two past weeks I have been trying to get Haskell working on my macbook pro with Snow Leopard, but unfortunately I was unable to get it. There are some solutions on the internet, but these are mostly useful when you already have ghci and haskell platform installed on your computer, while I can't even get it installed. I have Xcode 3.1 installed, but it seems that it is not the case. Maybe you could recommend any solution to this problem? Sincerely, Bernardas Jankauskas P.S.: I have attached two screenshots of what occurs for me when I am trying to install haskell platform and ghci.

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Re: \x -> x < 0.5 && x > -0.5
by Stephen Tetley 20 Oct '09

20 Oct '09

Christian Maeder wrote: > Hoogle did not find a function of type: > > (b -> b -> b) -> (a -> b) -> (a -> b) -> a -> b > > or (b -> c -> d) -> (a -> b) -> (a -> c) -> a -> d > > But maybe such a function is worth being added to Data.Function. Hello Christian This is the big Phi or S' combinator. I would certainly second its addition to Data.Function. sprime p q r s = p (q s) (r s) Best wishes Stephen

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new locale?
by Robert Wills 19 Oct '09

19 Oct '09

Hello, http://hackage.haskell.org/package/old-locale-1.0.0.1 says "This package provides the old locale library. For new code, the new locale library is recommended." Does anyone know where I would find the new library? Thanks, Rob

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parsing upto n items with parsec
by Ashish Agarwal 19 Oct '09

19 Oct '09

Hi. I'm just learning Parsec and Haskell. It is a great library, and I see how "many" lets you parse 0 or more items, "many1" parses 1 or more items, and "count" parses exactly n items. However, there is no combinator that parses between m and n items. What would be the best implementation for this?

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