Travis, because 'speedTest (read $ head args) `seq` return ()' is semantically equivalent to 'return ()' whose type is IO () therefore 'v <- a' assigns () to v. 'time (return $! speedTest (read $ head args))' is better... however, it's still not going to do what you want (time how long it takes speedTest to evaluate, I'm assuming), because (speedTest (read $ head args)) will only get evaluated once. The only way to actually time how long speedTest takes is to pass a function e.g. (Int -> IO t) to 'time', and have 'time' pass it a dummy argument, so that GHC can't optimize the evaluation of speedTest out. Steve Travis Erdman wrote:

Why oh why doesn't this work as I expect, and what is the simplest fix?

speedTest 4 = 44

time :: (Show t) => IO t -> IO t time a = do start <- getCPUTime v <- a end <- getCPUTime let diff = (fromIntegral (end - start)) / (10^12) print v -- WHY DOESNT THIS WORK? ALWAYS PRINTS () .... printf "Computation time: %0.3f sec\n" (diff :: Double) return v

main :: IO () main = do args <- getArgs time $ speedTest (read $ head args) `seq` return ()

------------------------------------------------------------------------

_______________________________________________ Beginners mailing list Beginners@haskell.org http://www.haskell.org/mailman/listinfo/beginners