paolo.veronelli:
Hi, reading a previous thread I got interested. I simplified the example pointed by dons in
import Control.Parallel
main = a `par` b `pseq` print (a + b ) where a = ack 3 11 b = ack 3 11
ack 0 n = n+1 ack m 0 = ack (m-1) 1 ack m n = ack (m-1) (ack m (n-1))
compiled with ghc --make prova -O2 -threaded
timings paolino@paolino-casa:~$ time ./prova +RTS -N1 32762
real 0m7.031s user 0m6.304s sys 0m0.004s paolino@paolino-casa:~$ time ./prova +RTS -N2 32762
real 0m6.997s user 0m6.728s sys 0m0.020s paolino@paolino-casa:~$
without optimizations it gets worse
paolino@paolino-casa:~$ time ./prova +RTS -N1 32762
real 1m20.706s user 1m18.197s sys 0m0.104s paolino@paolino-casa:~$ time ./prova +RTS -N2 32762
real 1m38.927s user 1m45.039s sys 0m0.536s paolino@paolino-casa:~$
staring at the resource usage graph I can see it does use 2 cores when told to do it, but with -N1 the used cpu goes 100% and with -N2 they both run just over 50%
thanks for comments
Firstly, a and b are identical, so GHC commons them up. The compiler transforms it into: a `par` a `seq` print (a + a) So you essentially fork a spark to evaluate 'a', and then have the main thread also evaluate 'a' again. One of them wins, then you add the result to itself. The runtime may choose not to convert your first spark into a thread. Running with a 2009 GHC head snapshot, we can see with +RTS -sstderr SPARKS: 1 (0 converted, 0 pruned) That indeed, it doesn't convert your `par` into a real thread. While, for example, the helloworld on the wiki: http://haskell.org/haskellwiki/Haskell_in_5_steps Converts 2 sparks to 2 theads: SPARKS: 2 (2 converted, 0 pruned) ./B +RTS -threaded -N2 -sstderr 2.13s user 0.04s system 137% cpu 1.570 total -- Don