On 25 June 2010 10:57, Frank Moore
Hello Haskellers,
I am new to programming in Haskell and I am having trouble understanding exactly when statements become evaluated. My goal is to try and measure how long a computation takes without having to use a show function. The code I am trying to use is below (taken in part from RWH chapter 25)
---------------------------------- import Data.List (foldl') import Data.Time.Clock (diffUTCTime, getCurrentTime) import Control.DeepSeq (deepseq)
mean :: [Double] -> Double mean xs = s / fromIntegral n where (n,s) = foldl' k (0,0) xs k (n,s) x = n `seq` s `seq` (n+1,s+x)
main = do let as = [1..1e7] :: [Double] start <- getCurrentTime let meanOver2 = deepseq (mean as) `seq` (mean as) / fromIntegral 2 end <- getCurrentTime putStrLn (show (end `diffUTCTime` start)) putStrLn (show meanOver2) -------------------------------------
My understanding of deepseq was that it evaluates (mean as) completely before continuing, and then the show would not take any time, but instead all the time is spent in the show meanOver2 function. I feel like I am missing something fundamental here. Any suggestions? Thanks for your help.
It does... but because you don't save the result it doesn't keep the result (you're wanting Common Sub-expression Elimination, which GHC doesn't do). Try this: let meanAs = mean as meanOver2 = meanAs `deepSeq` meanAs / 2 Note that your usage of "fromIntegral" isn't required, as that is automatically done for any integral literal (but if you had 2 in an Int variable then you would need fromIntegral). -- Ivan Lazar Miljenovic Ivan.Miljenovic@gmail.com IvanMiljenovic.wordpress.com