Re: [Haskell-cafe] 3rd party widgets with qtHaskell (Marble)
On Wed, 2010-03-10 at 11:22 +0100, Alp Mestanogullari wrote: This something you are afaik able to do. I'm cc'ing David (qthaskell's author). Thanks for the reply. I've worked it out. The below code demonstrates getting and setting a property from a marble widget. I'm a little surprised it worked. If my C++ is right what I've done here is dynamically cast the Marble widget as it's Qt parent. This of course is fine, but given that longitude() is not a virtual function on the parent, I'd to have to cast as the child to access this function - my code shouldn't have scope of Marble specific functions. This doesn't follow C++ (or I need to brush up on my OO programming!). I can live with this, after all I'm not writing a C++ program, but if anyone can explain this I'd be interested to understand why. One other peculiarity I noticed was that the qVariant constructor will only take Double or Integer types if they are nested in a tuple. Again, this is fine, but at odds with the documentation which implies constructors can take: () | p1 | (p1) | (p1,p2,...pn) So I'm over the first hurdle; it is possible, now to think of something interesting to do with it :-) module Main where import Qtc main :: IO () main = do app - qApplication () rok - registerResource marble.rcc loader - qUiLoader () uiFile - qFile :/marble.ui open uiFile fReadOnly ui - load loader uiFile close uiFile () ui_map - findChild ui (QWidget*, marbleWidget) ui_button - findChild ui (QPushButton*, pushButton) sc - qObjectProperty ui_map longitude bt - qObjectProperty ui_button text long - qVariantValue_Double sc print long x - qVariant (-1::Double) blah - qObjectSetProperty ui_map longitude x qshow ui () ok - qApplicationExec () return () ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] 3rd party widgets with qtHaskell (Marble)
Hi, I know this isn't a qtHaskell list, but I don't think there is one. Was wondering if anyone has any ideas on the below. Basically I'm trying to control a Marble (Map software) Qt widget from qtHaskell. So I've mocked up a very simple user interface in Qt Designer (1 form, 1 Marble widget). I can load this up and display it fine in Haskell, but as soon as I try to interrogate the widget I get a seg fault (eg qObjectProperty) My guess is that the call to findChild, although it executes OK it is not producing a valid QObject - probably casting to Marble::MarbleWidget* it crux of the problem. I can get this working using standard Qt Widgets (just like the examples show from qtHaskell), so I know the method is sound - although calling 3rd party widgets like this may be ambitious or impossible. I recognise this is a fairly broad query! Has anyone tried anything similar? Is it even possible to do this in qtHaskell as I'm proposing? I'm a Qt novice, so it may well be that I've misunderstood qtHaskell. Cheers, Phil. Using: GHC 6.12.1 / QT4.5 / Marble 0.8 / Ubuntu 9.04 module Main where import Qtc main :: IO () main = do app - qApplication () rok - registerResource marble.rcc loader - qUiLoader () uiFile - qFile :/marble.ui open uiFile fReadOnly ui - load loader uiFile close uiFile () ui_map - findChild ui (Marble::MarbleWidget*, MarbleWidget) sc - qObjectProperty ui_map showCompass qshow ui () ok - qApplicationExec () return () ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] Re: Parallel foldl doesn't work correctly
If you still have trouble, then try using ThreadScope http://code.haskell.org/ThreadScope/ with GHC 6.12.1. You can use ThreadScope directly from the darcs repository on code.haskell.org, and we hope to do a proper release soon. Cheers, Simon Thanks for the advice, just downloaded ThreadScope and it's pretty useful (before I was using Ubuntu's System Monitor which isn't ideal). I've moved onto 6.12 and I now have my program working nicely over 2 cores - the problem was at least in part my own design - I was generating large thunks in my parallel version which was killing performance. With this solved 2 cores gives me ~50% performance increase. What I'm doing now is taking a list I am going to fold over and splitting it up so I have a list of lists, where each parent list element representing work for 1 core. I then fold lazily and only parallelise on the final sum operation which (as far as I can see) sends each chunk of folds to a different core and sums the results. Can I confirm - what you are suggesting is that although I can't parallelise fold itself, I could force evaluation on the list I am about to fold in parallel and then merely accumulate the result at the end -- thus most the donkey work is done in parallel? If this is possible, it may be more flexible then my method. I'm lucky as each fold operation will take give-or-take the same amount of time so I can just chunk up fold jobs for each core equally. If this wasn't the case (and it certainly won't always be!), parallelising on individual items would be the way to go. Thanks, Phil. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Re: Parallel foldl doesn't work correctly
On Sat, 2009-12-12 at 13:46 +, Ben Millwood wrote: On Sat, Dec 12, 2009 at 10:08 AM, Maciej Piechotka uzytkown...@gmail.com wrote: If operation is associative it can be done using divide et impera spliting list in half and operating on it pararerlly then split in half etc. Thank you very much for the replies. I've come to the conclusion that, yep, you can't (directly) parallelise of fold operation, as fold guarantees order of processing. With something like map the runtime is free to continue sparking function application to each element without waiting for the result. So we spark f x, force evaluation of the remainder of the xs and recurse. I'm *guessing* at a detailed level when we are creating the output list, haskell can concat each result element before f x returns due to laziness - that is, haskell doesn't need to wait for evaluation of f x, before continuing? With fold, and specifically with foldl (+), this isn't the case as (+) is strict on both arguments and thus it cannot continue until each sparked evaluation has completed and combined with the accumulator. If (+) was not strict on both arguments, I'm not sure if could solider on... assuming I've understood map correctly!? Writing it out long hand (sorry if this is tedious!), we have: using :: a - Strategy a - a using x s = s x `seq` x rwhnf :: Strategy a rwhnf x = x `seq` () parList :: Strategy a - Strategy [a] parList strat [] = () parList strat (x:xs) = strat x `par` (parList strat xs) parMap :: Strategy b - (a - b) - [a] - [b] parMap strat f = (`using` parList strat) . map f 'using' applies a strategy to an item, and then returns the item. 'parList' is a (combinator) strategy which applies an atomic strategy to each element in the list *in parallel* (for example forcing each element to WHNF). So for parMap we have xs passed into 'map f' - the result is then passed to 'using' which will force application of 'f' on each element in parallel by way of 'parList'. No forced evaluation is dependant on a previous evaluation. Now for parFoldl - a crude and wrong representation for my purposes could be: parFoldl :: Num b = Strategy b - (a - b) - [a] - b parFoldl strat f = sum . (`using` parList strat) . map f This isn't really a fold of course, but it is doing roughly the same thing, it's summing the results of applying function 'f' to each element in a list. The problem here is that sum will only allow one spark at a time, because sum [] = 0 sum (x:xs) = x + sum xs So we get something like: 0 + (x4 + (x3 + (x2 + (x1 For example the result for (x4 + previous) can only be evaluated after x3, x2 and x1 have been evaluated. This means it won't spark evaluation on x4 until (x3 + ) has been evaluated, thus only one core is ever used. I believe fold is just the general case of sum and the same logic applies. I suppose my questions are: Have I got this right, if not very succinct!? Is it purely the strictness of (+) that causes this situation? Ignoring DPH, is it possible to write a parallel fold avoiding something like the technique below? Anyhow, a workaround similar to those suggested I came up with is to divide the folds up across the cores and then sum the sub-folds - this produces approximately double the performance across two cores: import Control.Parallel.Strategies (parMap,rwhnf) import Data.List (foldl') import Data.List.Split (chunk) import GHC.Conc (numCapabilities) -- Prepare to share work to be -- done across available cores chunkOnCpu :: [a] - [[a]] chunkOnCpu xs = chunk (length xs `div` numCapabilities) xs -- Spark a fold of each chunk and -- sum the results. Only works because -- for associative folds. foldChunks :: ([a] - a) - (a - b - a) - a - [[b]] - a foldChunks combineFunc foldFunc acc = combineFunc . (parMap rwhnf $ foldl' foldFunc acc) -- Some pointless work to keep thread busy workFunc :: Int - Int workFunc 1 = 1 workFunc x = workFunc $ x - 1 -- Do some work on element x and append foldFunc :: Int - Int - Int foldFunc acc x = acc + workFunc x testList = repeat 10 answer = foldChunks sum foldFunc 0 $ chunkOnCpu (take 50 testList) main :: IO() main = print answer ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Re: Parallel foldl doesn't work correctly
-- Prepare to share work to be -- done across available cores chunkOnCpu :: [a] - [[a]] chunkOnCpu xs = chunk (length xs `div` numCapabilities) xs -- Spark a fold of each chunk and -- sum the results. Only works because -- for associative folds. foldChunks :: ([a] - a) - (a - b - a) - a - [[b]] - a foldChunks combineFunc foldFunc acc = combineFunc . (parMap rwhnf $ foldl' foldFunc acc) I should probably point out that use of chunk above isn't a good idea in anything beyond a toy example. If you have used a list comprehension to create your input then splitting it like the above results in thunks that grow with list size as chunk forces generation of the list. This rapidly negates any advantage gained from processing across 1 core! This is easily solved - just alter the generating function to create a *list* of list comprehensions equal in length to the number of cores you wish to process across, rather than create one list that is split across the cores later. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] Parallel foldl doesn't work correctly
Hi, Can anyone put me right here. I am trying to use a setup similar to parMap to spark each valuation in a list in parallel, where the resulting (evaluated) list is folded to produce a final single result. Having done the obligatory google, I modified a few common examples to give: pfoldl f acc xs = foldl' f acc (xs `using` parList rwhnf) This compiles and if I monitor my CPUs it starts to use both cores, but after approx 10 seconds, one core drops to low rate (I'm using a 2 core machine). The end result is that -N2 is actually a bit slower than -N1! I'm guessing I haven't grasped the concept properly - although as map is just 'foldl (+) 0' I'm at a loss to see why this approach wouldn't work given it is pretty similar to parMap - could anyone point out what I'm missing? If it's any use the context of the code is below. Many thanks! Phil. mc :: MonteCarloUserData - [[Double]] - Double mc userData rndss = existentialResult (pfoldl f existenAvg rndss) $ numOfSims userData where f = flip $ existentialCombine . payOff' . expiryValue payOff' = existentialPayOff userData expiryValue = foldl' (existentialEvolve userData) (stock userData) existenAvg = averager userData ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
