You're right, without the show, no work is done.
But I'm puzzled. I thought waitAny would have caused one task to be executed.
If that doesn't wait for the async to compute a value (and not some
thunk) I don't see how to use asyncs, so I'm obviously missing
something.
How can I force deeper evaluation?
[in the end p would be a full parser, so whatever it is needed to
cause the parsing needs to be done outside of it]
On Fri, Apr 24, 2015 at 10:44 AM,
On 24/04/2015 14:21, Maurizio Vitale wrote:
G'day, I have a test code that I compile with ghc -threaded -eventlog -rtsopts --make parallel.hs and run with ./parallel 2 +RTS -ls -N4 on a laptop with 4 physical cores. I would expect activities in two threads, but threadscope shows only one active thread. Can somebody explain me the program's behaviour?
Thanks a lot
Maurizio
{-# LANGUAGE UnicodeSyntax #-} {-# LANGUAGE TupleSections #-}
import Control.Applicative import Control.Concurrent.Async (async, waitAny, Async) import Data.List (delete, sortBy) import Data.Ord (comparing) import System.CPUTime import System.Environment import GHC.Conc (numCapabilities)
concurrentlyLimited :: Int -> [IO a] -> IO [a] concurrentlyLimited n tasks = concurrentlyLimited' n (zip [0..] tasks) [] []
concurrentlyLimited' ∷ Int -- ^ number of concurrent evaluations → [(Int, IO b)] -- ^ still to run (ordered by first element) → [Async (Int,b)] -- ^ currently running → [(Int,b)] -- ^ partial results (ordered by first element) → IO [b] concurrentlyLimited' _ [] [] results = return . map snd $ sortBy (comparing fst) results concurrentlyLimited' 0 todo ongoing results = do (task, newResult) <- waitAny ongoing concurrentlyLimited' 1 todo (delete task ongoing) (newResult:results)
concurrentlyLimited' _ [] ongoing results = concurrentlyLimited' 0 [] ongoing results concurrentlyLimited' n ((i, task):otherTasks) ongoing results = do t <- async $ (i,) <$> task concurrentlyLimited' (n-1) otherTasks (t:ongoing) results
euler :: Int → Int euler n = length (filter (relprime n) [1..n-1]) where relprime x y = gcd x y == 1
sumEuler :: Int → Int sumEuler = sum . (map euler) . mkList where mkList n = [1..n-1]
p ∷ IO Int p = return $ sumEuler 3000
numThreads ∷ [String] → IO Int numThreads [] = return numCapabilities numThreads [cores] = return $ read cores
main ∷ IO() main = do threads ← getArgs >>= numThreads putStrLn $ "Running up to " ++ show threads ++ " threads in parallel (on " ++ show numCapabilities ++ " cores)" startTime ← getCPUTime f ← concurrentlyLimited threads $ replicate 10 p endTime ← getCPUTime putStrLn $ foldr ((++) . show ) "" f putStrLn $ "Evaluation took " ++ show (fromIntegral (endTime - startTime) / 1000000000000∷Double) _______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
Hello,
I don't have an Haskell environnement at hand, but my advice is to search for when your Async/eulerSum calls are evaluated. For example try to remove this line -- putStrLn $ foldr ((++) . show ) "" f Does your program still compute something ? If no that's because your sum is evaluated due to the show and not due to your async.
t <- async $ (i,) <$> task
Your async will try to compute (i,eulerSum) but you never force the computation of the eulersum inside the async, so the async take no time and return quickly. Instead of this type [Async (Int, b)] you should aim for this one [(Int, Async b)]
Let me know if that helps you.
Regards, Romain
_______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
On Fri, Apr 24, 2015 at 1:44 PM,
On 24/04/2015 14:21, Maurizio Vitale wrote:
G'day, I have a test code that I compile with ghc -threaded -eventlog -rtsopts --make parallel.hs and run with ./parallel 2 +RTS -ls -N4 on a laptop with 4 physical cores. I would expect activities in two threads, but threadscope shows only one active thread. Can somebody explain me the program's behaviour?
Thanks a lot
Maurizio
{-# LANGUAGE UnicodeSyntax #-} {-# LANGUAGE TupleSections #-}
import Control.Applicative import Control.Concurrent.Async (async, waitAny, Async) import Data.List (delete, sortBy) import Data.Ord (comparing) import System.CPUTime import System.Environment import GHC.Conc (numCapabilities)
concurrentlyLimited :: Int -> [IO a] -> IO [a] concurrentlyLimited n tasks = concurrentlyLimited' n (zip [0..] tasks) [] []
concurrentlyLimited' ∷ Int -- ^ number of concurrent evaluations → [(Int, IO b)] -- ^ still to run (ordered by first element) → [Async (Int,b)] -- ^ currently running → [(Int,b)] -- ^ partial results (ordered by first element) → IO [b] concurrentlyLimited' _ [] [] results = return . map snd $ sortBy (comparing fst) results concurrentlyLimited' 0 todo ongoing results = do (task, newResult) <- waitAny ongoing concurrentlyLimited' 1 todo (delete task ongoing) (newResult:results)
concurrentlyLimited' _ [] ongoing results = concurrentlyLimited' 0 [] ongoing results concurrentlyLimited' n ((i, task):otherTasks) ongoing results = do t <- async $ (i,) <$> task concurrentlyLimited' (n-1) otherTasks (t:ongoing) results
euler :: Int → Int euler n = length (filter (relprime n) [1..n-1]) where relprime x y = gcd x y == 1
sumEuler :: Int → Int sumEuler = sum . (map euler) . mkList where mkList n = [1..n-1]
p ∷ IO Int p = return $ sumEuler 3000
numThreads ∷ [String] → IO Int numThreads [] = return numCapabilities numThreads [cores] = return $ read cores
main ∷ IO() main = do threads ← getArgs >>= numThreads putStrLn $ "Running up to " ++ show threads ++ " threads in parallel (on " ++ show numCapabilities ++ " cores)" startTime ← getCPUTime f ← concurrentlyLimited threads $ replicate 10 p endTime ← getCPUTime putStrLn $ foldr ((++) . show ) "" f putStrLn $ "Evaluation took " ++ show (fromIntegral (endTime - startTime) / 1000000000000∷Double) _______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
Hello,
I don't have an Haskell environnement at hand, but my advice is to search for when your Async/eulerSum calls are evaluated. For example try to remove this line -- putStrLn $ foldr ((++) . show ) "" f Does your program still compute something ? If no that's because your sum is evaluated due to the show and not due to your async.
t <- async $ (i,) <$> task
Your async will try to compute (i,eulerSum) but you never force the computation of the eulersum inside the async, so the async take no time and return quickly. Instead of this type [Async (Int, b)] you should aim for this one [(Int, Async b)]
Let me know if that helps you.
Regards, Romain
_______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners