It's important to switch from mod to rem. This can be done by a simple abstract interpretation. I'm nore sure if it's jhc or gcc that does this for jhc.
-- Lennart On Sat, Mar 27, 2010 at 10:30 PM, Rafael Cunha de Almeida <almeida...@gmail.com> wrote: > John Meacham wrote: >> Here are jhc's timings for the same programs on my machine. gcc and ghc >> both used -O3 and jhc had its full standard optimizations turned on. >> >> jhc: >> ./hs.out 5.12s user 0.07s system 96% cpu 5.380 total >> >> gcc: >> ./a.out 5.58s user 0.00s system 97% cpu 5.710 total >> >> ghc: >> ./try 31.11s user 0.00s system 96% cpu 32.200 total >> >> >> As you can see, jhc shines at this example, actually beating gcc -O3. It >> isn't too surprising, this is exactly the sort of haskell code that jhc >> excels at. > > What's the property of that code which makes jhc excels in it? What > makes ghc perform so poorly in comparison? > _______________________________________________ > Haskell-Cafe mailing list > Haskell-Cafe@haskell.org > http://www.haskell.org/mailman/listinfo/haskell-cafe > _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
