stefanor:
On Wed, Jun 20, 2007 at 01:24:00PM +1000, Donald Bruce Stewart wrote:
aeyakovenko:
$ time ./md5sum ./md5sum [105,252,52,138,187,192,216,17,225,123,185,3,124,101,86,132]
real 0m4.790s user 0m3.688s sys 0m0.492s
$ time md5sum ./md5sum 69fc348abbc0d811e17bb9037c655684 ./md5sum
real 0m0.023s user 0m0.000s sys 0m0.008s
this is my implementation using crypto (http://www.haskell.org/crypto/). Am I doing something wrong?
module Main where
import System import qualified Data.Digest.MD5 as MD5 import qualified Data.ByteString as BS
main = do args <- getArgs dt <- BS.readFile $ head args putStrLn $ show $ MD5.hash . BS.unpack $ dt ^^^^ ^^^^^^^^^
not a good idea.
You need an MD5 over bytestrings, not [Word8].
Wouldn't deforestation have produced something within a factor of 4-ish of optimal?
There's no deforestation between ByteString.unpack and MD5.hash. So all the time here is spent taking some big bytestring, and unpacking it into n list cons cells. md5ing the bytestring directly should knock 95% off that. -- Don