Thanks for all the advises so far. Ok, here's my monster that need to be timed in order to make a comparison: (it's about the control digit of SEDOL numbers http://en.wikipedia.org/wiki/SEDOL ): knip _ [] = Nothing knip k xs = Just (splitAt k xs) ip xs = sum . zipWith (*) xs an = ['0'..'9']++['A'..'Z'] s = take 841 $ cycle "0987654321" f = \xs -> xs ++ [(sna!!).ip [1,3,1,7,3,9]. map (flip (fromJust .) an . findIndex . (==))$xs] Here's my try for timing: *Main> (foldl1 (\x y -> f y) .concat.replicate 1000000 $ unfoldr (knip 6) an) "UVWXYZ7" (1.31 secs, 330291000 bytes) (It's incl. the construction of the test list, as is in the language to compare ) I need the whole list to be evaluated. Interpreted mode IS A MUST :-) BTW I increased stack size thanks Don Stewart schreef:
bradypus:
Suppose I've:
f = map g
I want to know how much time it takes (interpreted mode) to fully process list xs (at least 1e6 elements) with function g. Is it sufficient to execute:
*Main> last . f $ xs <result> (x.xx secs, yyyyyyyyyyy bytes)
Are there any hidden difficulties involved?
Reason is: comparing timings Haskell vs an interpreted language without laziness.
If you care about timings, it's probably a better idea to compile the code (with optimisations on), to get a better idea of what the code would do in a production environment.
You could then just time the binary,
main = print . sum $ ....
ghc -O2 A.hs --make time ./A
-- Don