May 2009 - Beginners - Haskell.org

Re: [Haskell-beginners] How to wait till a process is finished before invoking the next one?
by Daniel Fischer 07 May '09

07 May '09

Am Donnerstag 07 Mai 2009 20:53:11 schrieb Thomas Friedrich: > Hi Daniel and everyone, > > Thanks for the reply! > > I thought of using waitForProcess, and in fact an earlier version of the > program did. However, as the program got more complex, I don't really > see how this is still possible. > > For example the runProgram function looks like this: > > runProgram :: [String] -> IO () > runProgram [] = return () > runProgram (c:cs) = do > runCommand ("lalala " ++ c) > runProgram cs You could have runCommand ("lalala " ++ c) >>= waitForProcess in the penultimate line. That would ensure that the command has finished before its results are requested, but of course destroy all possibilities of parallelism :-( Another option would be runProgram [] = return [] runProgram (c:cs) = unsafeInterleaveIO $ do ph <- runCommand ("lalala " ++ c) phs <- runProgram cs return (ph:phs) and then, processHandles <- runProgram blah mapM_ waitForProcess (processHandles) next step that should work (I hope) and would allow the commands to be run in parallel while making sure all have finished before the next step is started. > > It might be possible to write the function runProgram in a way, so that > it returns an expression of type [IO ProcessHandle] and then try to work > from there. But I have the feeling, that this will become messy very > quickly, and there must be some more elegant way of doing this. The > other thing is that actually not all functions are able to return > ProcessHandles, e.g. > > writeFeatures :: [String] -> IO () > writeFeatures cs = Exc.bracket (openFile training AppendMode) hClose (\h > -> goo h) > where > goo h = go 1 cs > where > go :: Int -> [String] -> IO () > go n [] = putStrLn "Features written." > go n (c:cs) = do > features <- makeFeatures n c -- makeFeatures :: Int -> > String -> IO String > hPutStr h features > go (n+1) cs > > And the file that is produced here is needed in the next function. That can't appear in runProgram, though. I'm not sure it would work, but you coud give {-# LANGUAGE BangPatterns #-} ... !a <- writeFeatures blah more or go n (c:cs) = do features <- makeFeatures n c !a <- hPutStr h features go (n+1) cs a try. > > I hoped to do something with forkIO, as I would like to parallelize the > whole program at the end. Especially the function runProgram would > benefit hugely from this (I so don't have a clue how to do this yet;). I > tried for example the following: > > main :: IO () > main = do > cs <- getArgs > p1 <- forkIO $ writeData cs > p2 <- forkIO $ runProgram cs > p3 <- forkIO $ writeFeatures cs > p4 <- forkIO $ runTestOnFeatures > seq p1 (seq p2 (seq p3 (seq p4 (putStrLn "Done")))) > > But that of course doesn't work, because now I am not actually > requesting anything. The program does in fact nothing, apart from > printing out "Done". Huh. There doesn't seem to be a wait function for ThreadIds, so I guess you would have to communicate via MVars, QSemNs or some such means to signal that one task has been completed and the next can be started. > > Any ideas? > > Cheers, > Thomas

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parallel program in haskell in 5 steps
by Jack Kennedy 06 May '09

06 May '09

In step 4 of Haskell in 5 Steps [ http://haskell.org/haskellwiki/Haskell_in_5_steps] a parallel program is given. I changed it very slightly so it would run a long time (see below). It compiles and runs but my CPU meter is barely above 50%. I have a dual core processor. What in the world would keep this program from completely saturating the CPU? import Control.Parallel main = a `par` b `pseq` print (a + b) where a = ack 4 10 b = ack 4 10 fac 0 = 1 fac n = n * fac (n-1) ack 0 n = n+1 ack m 0 = ack (m-1) 1 ack m n = ack (m-1) (ack m (n-1)) fib 0 = 0 fib 1 = 1 fib n = fib (n-1) + fib (n-2)

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Context reducion Stack overflow
by Marco Túlio Gontijo e Silva 06 May '09

06 May '09

Hello, I'm trying to compile the following code: > import Data.IORef > class F f where > m :: (a -> a) -> f a -> f a > class M m a where > mm :: (a -> a) -> m -> IO () > instance (F f, M m (f a)) => M m a where > mm f v = mm (m f) v And I'm getting: $ ghci teste.lhs -XMultiParamTypeClasses -XFlexibleInstances -XFlexibleContexts -XUndecidableInstances -XIncoherentInstances GHCi, version 6.10.1: http://www.haskell.org/ghc/ :? for help Loading package ghc-prim ... linking ... done. Loading package integer ... linking ... done. Loading package base ... linking ... done. Failed, modules loaded: none. Prelude> ^DLeaving GHCi. GHCi, version 6.10.1: http://www.haskell.org/ghc/ :? for help Loading package ghc-prim ... linking ... done. Loading package integer ... linking ... done. Loading package base ... linking ... done. [1 of 1] Compiling Main ( teste.lhs, interpreted ) teste.lhs:1:0: Context reduction stack overflow; size = 20 Use -fcontext-stack=N to increase stack size to N `$dM{atp} :: {M m (f (f (f (f (f (f (f (f (f (f (f (f (f (f (f (f (f (f (f (f a))))))))))))))))))))}' arising from a use of `mm' at teste.lhs:10:13-22 `$dM{atm} :: {M m (f (f (f (f (f (f (f (f (f (f (f (f (f (f (f (f (f (f (f a)))))))))))))))))))}' arising from a use of `mm' at teste.lhs:10:13-22 `$dM{atj} :: {M m (f (f (f (f (f (f (f (f (f (f (f (f (f (f (f (f (f (f a))))))))))))))))))}' arising from a use of `mm' at teste.lhs:10:13-22 `$dM{atg} :: {M m (f (f (f (f (f (f (f (f (f (f (f (f (f (f (f (f (f a)))))))))))))))))}' arising from a use of `mm' at teste.lhs:10:13-22 `$dM{atd} :: {M m (f (f (f (f (f (f (f (f (f (f (f (f (f (f (f (f a))))))))))))))))}' arising from a use of `mm' at teste.lhs:10:13-22 `$dM{ata} :: {M m (f (f (f (f (f (f (f (f (f (f (f (f (f (f (f a)))))))))))))))}' arising from a use of `mm' at teste.lhs:10:13-22 `$dM{at7} :: {M m (f (f (f (f (f (f (f (f (f (f (f (f (f (f a))))))))))))))}' arising from a use of `mm' at teste.lhs:10:13-22 `$dM{at4} :: {M m (f (f (f (f (f (f (f (f (f (f (f (f (f a)))))))))))))}' arising from a use of `mm' at teste.lhs:10:13-22 `$dM{at1} :: {M m (f (f (f (f (f (f (f (f (f (f (f (f a))))))))))))}' arising from a use of `mm' at teste.lhs:10:13-22 `$dM{asY} :: {M m (f (f (f (f (f (f (f (f (f (f (f a)))))))))))}' arising from a use of `mm' at teste.lhs:10:13-22 `$dM{asV} :: {M m (f (f (f (f (f (f (f (f (f (f a))))))))))}' arising from a use of `mm' at teste.lhs:10:13-22 `$dM{asS} :: {M m (f (f (f (f (f (f (f (f (f a)))))))))}' arising from a use of `mm' at teste.lhs:10:13-22 `$dM{asP} :: {M m (f (f (f (f (f (f (f (f a))))))))}' arising from a use of `mm' at teste.lhs:10:13-22 `$dM{asM} :: {M m (f (f (f (f (f (f (f a)))))))}' arising from a use of `mm' at teste.lhs:10:13-22 `$dM{asJ} :: {M m (f (f (f (f (f (f a))))))}' arising from a use of `mm' at teste.lhs:10:13-22 `$dM{asG} :: {M m (f (f (f (f (f a)))))}' arising from a use of `mm' at teste.lhs:10:13-22 `$dM{asD} :: {M m (f (f (f (f a))))}' arising from a use of `mm' at teste.lhs:10:13-22 `$dM{asA} :: {M m (f (f (f a)))}' arising from a use of `mm' at teste.lhs:10:13-22 `$dM{asx} :: {M m (f (f a))}' arising from a use of `mm' at teste.lhs:10:13-22 `$dM{asj} :: {M m (f a)}' arising from a use of `mm' at teste.lhs:10:13-22 Failed, modules loaded: none. An example of instances of these classes: > instance F [] where > m = map > instance M (IORef a) a where > mm = flip modifyIORef I don't get what's the problem with my definition. Can someone clarify me on this? Greetings. -- marcot http://marcot.iaaeee.org/

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question >>= [style, efficiency]
by Joel Neely 06 May '09

06 May '09

I can think of a variety of ways to put a value at the end of a list of the same type. -- cutesy-clever putAtEnd :: [a] -> a -> [a] putAtEnd xs x = reverse (x : reverse xs) -- by hand atEnd :: [a] -> a -> [a] atEnd [] z = [z] atEnd (x:xs) z = x : atEnd xs z -- obvious brute force infixr 9 +: (+:) :: [a] -> a -> [a] (+:) xs x = xs ++ [x] I'd appreciate feedback/advice on whether: 1) I've missed any clearly-better alternatives; 2) any of those is obviously better or worse than the others; 3) I should have known where to look for the answer myself (instead of brute-force Googling). I've semi-inadvertently mixed two issues above: infix-vs-prefix and algorithm design. TIA for suggestions on either. -jn- -- Beauty of style and harmony and grace and good rhythm depend on simplicity. - Plato

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haskell IO help
by Ashley Banks 06 May '09

06 May '09

From: applebiz89(a)hotmail.com To: beginners(a)haskell.org Subject: Date: Wed, 6 May 2009 13:28:18 +0000 I havent done much IO at all in haskell, only within the function itself. However I want to get the input from the interface for the function and havent done this before. // Say I have this as my data type and list of films data Film = Film String String Int [String] -- List of films testDatabase :: [Film] testDatabase = [(Film "Casino Royale" "Martin Campbell" 2006 ["Garry", "Dave", "Zoe"]) ] // with functions such as: becomeFan :: String -> String -> [Film] -> [Film] becomeFan _ _ [] = [] becomeFan filmName fanName ((Film title director year fans):xs) | filmName == title = (Film title director year fanName:fans) : xs | otherwise = (Film title director year fans) : becomeFan filmName fanName xs filmsInGivenYear :: Int -> [Film] -> [String] filmsInGivenYear filmYear films = [ title | (Film title director year fans) <- films, year == filmYear] // I want to ask the user what function they want to use, I have this so far main :: IO() do putStr "Hi there! what is your name: " fanName = getLine do putStr "1 = Insert film, 2 = Become a Fan, 3 = The number of fans of a film, 4 = Film released in a year: " input = getLine read input :: Int (if input == 1 then main x = insertFilm [] else if input == 2 then main x = becomeFan [] else if input == 3 then main x = numberOfFans []) // I thought using the if statement would work, but now I cant think how to gather the needed input for the function they have chosen? thanks apple Get the New Internet Explore 8 Optimised for MSN. Download Now _________________________________________________________________ Share your photos with Windows Live Photos – Free. http://clk.atdmt.com/UKM/go/134665338/direct/01/

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(no subject)
by Ashley Banks 06 May '09

06 May '09

I havent done much IO at all in haskell, only within the function itself. However I want to get the input from the interface for the function and havent done this before. // Say I have this as my data type and list of films data Film = Film String String Int [String] -- List of films testDatabase :: [Film] testDatabase = [(Film "Casino Royale" "Martin Campbell" 2006 ["Garry", "Dave", "Zoe"]) ] // with functions such as: becomeFan :: String -> String -> [Film] -> [Film] becomeFan _ _ [] = [] becomeFan filmName fanName ((Film title director year fans):xs) | filmName == title = (Film title director year fanName:fans) : xs | otherwise = (Film title director year fans) : becomeFan filmName fanName xs filmsInGivenYear :: Int -> [Film] -> [String] filmsInGivenYear filmYear films = [ title | (Film title director year fans) <- films, year == filmYear] // I want to ask the user what function they want to use, I have this so far main :: IO() do putStr "Hi there! what is your name: " fanName = getLine do putStr "1 = Insert film, 2 = Become a Fan, 3 = The number of fans of a film, 4 = Film released in a year: " input = getLine read input :: Int (if input == 1 then main x = insertFilm [] else if input == 2 then main x = becomeFan [] else if input == 3 then main x = numberOfFans []) // I thought using the if statement would work, but now I cant think how to gather the needed input for the function they have chosen? thanks apple _________________________________________________________________ Beyond Hotmail — see what else you can do with Windows Live. http://clk.atdmt.com/UKM/go/134665375/direct/01/

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Re: Programação Funcional - TP1
by Marco Túlio Gontijo e Silva 05 May '09

05 May '09

Olá Henrique, seu problema Polepos.hs excedeu o tempo limite de execução no SPOJ. Algumas dicas: você pode escrever: cars = map fst startGrid positions = map snd startGrid (cars, positions) = unzip take (length positions) [0..] [0 .. pred $ length positions] do scoreBoard <- parseScoreboard n let startGrid = getStartGrid scoreBoard putStrLn $ showGrid startGrid main parseScoreboard n >>= putStrLn . showGrid . getStartGrid >> main showGrid Nothing = "-1" showGrid (Just xs) = foldl1 (++) $ intersperse " " $ map show xs showGrid = maybe "-1" (foldl1 (++) . intersperse " " . map show) Talvez fosse mais legal criar um tipo de dados e definir a classe Ord para eles do que usar tupla e sortBy. Essas dicas não estão relacionadas com a demora no tempo de execução. Isso se deve, acredito eu, ao seu algoritmo. -- marcot http://marcot.iaaeee.org/

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GHCI: Main function seems to mask other functions
by aditya siram 05 May '09

05 May '09

Hi all, I'm having a weird problem where when I add the 'main' function to my program and load it into GHCI, whenever I check the type or try to run my other functions I get a 'Not in scope: <functionname>' error. When I remove the 'main' function and reload the code, the problem goes away. I can't seem to reproduce this with a trivial example, but this has happened to me several times in my larger programs. thanks ... deech

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Polymorphism question from an OO-speaking newbie
by Joel Neely 04 May '09

04 May '09

Short version: Is it possible/reasonable to define a single function that accepts a single string or a list of strings (in which case it maps the single-string "flavor" over the list)? Longer version: Thus far I know that Haskell allows me to define a function on a single string, then define another function that maps the first one across a list of strings, as in: *Main> let quote1 s = "\"" ++ s ++ "\"" *Main> "Quux said " ++ quote1 "foo" ++ " loudly" "Quux said \"foo\" loudly" *Main> let quote = map quote1 *Main> quote ["foo", "baz", "bletch"] ["\"foo\"","\"baz\"","\"bletch\""] (BTW, is it standard terminology to say that quote lifts quote1 to lists?) In the above I have different names for the two functions. OO languages such as Java allow me to overload quote to accept either String or List<String> (and return the same type). AFAICT, Haskell's parametric polymorphism allows me to define functions (e.g. length) which deal with "listness" concepts indifferent to the type contained by the list. Am I missing something, or should I admit to OO wrongheadedness and accept that my inability to write a single declaration unifying: quote :: String -> String quote :: [String] -> [String] is an emphatic clue that I should change my expectations? Thanks in advance for any enlightenment shed my way! -jn- -- Beauty of style and harmony and grace and good rhythm depend on simplicity. - Plato

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Re: [Haskell-beginners] Parsec tutorial
by Keith Sheppard 02 May '09

02 May '09

Hi Patrick, This isn't what you asked for but it may be useful to you. I've created a parsec based SQL parser for my "TxtSushi" command-line program. It's still pretty early in development but it does parse a subset of the SELECT grammar. The project's page is http://keithsheppard.name/txt-sushi/index.html Keith

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