I don't know if this counts but how about
import Control.Applicative
import Control.Monad
import Random
import Data.List
main'' i j = replicateM j $ maximum' <$> (replicateM i . randomRIO $ (0,10^9))
maximum' = foldl1' max
t = main'' (10^4) 5
2009/10/9
Hi all,
I think there is something about my use of the IO monad that bites me, but I am bored of staring at the code, so here you g. The code goes through a list of records and collects the maximum in each record position.
-- test.hs import Random import System.Environment (getArgs) import System.IO (putStr)
samples :: Int -> Int -> IO [[Double]] samples i j = sequence . replicate i . sequence . replicate j $ randomRIO (0, 1000 ** 3)
maxima :: [[Double]] -> [Double] maxima samples@(_:_) = foldr (\ x y -> map (uncurry max) $ zip x y) (head samples) (tail samples)
main = do args <- getArgs x <- samples (read (head args)) 5 putStr . (++ "\n") . show $ maxima x
I would expect this to take constant memory (foldr as well as foldl), but this is what happens:
$ ghc -prof --make -O9 -o test test.hs [1 of 1] Compiling Main ( test.hs, test.o ) Linking test ... $ ./test 100 +RTS -p [9.881155955344708e8,9.910336352165401e8,9.71000686630374e8,9.968532576451201e8,9.996200333115692e8] $ grep 'total alloc' test.prof total alloc = 744,180 bytes (excludes profiling overheads) $ ./test 10000 +RTS -p [9.996199711457872e8,9.998928358545277e8,9.99960283632381e8,9.999707142123885e8,9.998952151508758e8] $ grep 'total alloc' test.prof total alloc = 64,777,692 bytes (excludes profiling overheads) $ ./test 1000000 +RTS -p Stack space overflow: current size 8388608 bytes. Use `+RTS -Ksize' to increase it. $
so...
does sequence somehow force the entire list of monads into evaluation before the head of the result list can be used? what can i do to implement this in constant memory?
thanks! matthias _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe