Repeated timing of function

Question

I have a very simple function f :: Int -> Int and I want to write a program that calls f for each n = 1,2,...,max. After each call of f the (cumulative) time that was used up to that point should be displayed (along with n and f n). How can this be implemented?

I'm still really new to input/output in Haskell, so this is what I've tried so far (using some toy example function f)

f :: Int -> Int
f n = sum [1..n]

evalAndTimeFirstN :: Int -> Int -> Int -> IO()
evalAndTimeFirstN n max time = 
  if n == max 
   then return () -- in the following we have to calculate the time difference from start to now
   else let str =  ("(" ++ (show n) ++  ", " ++ (show $ f n) ++ ", "++ (show time)++ ")\n") 
         in putStrLn str >> evalAndTimeFirstN (n+1) max time -- here we have to calculate the time difference

main :: IO()
main = evalAndTimeFirstN 1 5 0

I don't quite see how I have to introduce the timing here. (The Int for time probably has to be replaced with something else.)

Answer 1

I finally managed to find a solution. In this case we're measuring the "real" time in ms.

import Data.Time
import Data.Time.Clock.POSIX

f n = sum[0..n]

getTime = getCurrentTime >>= pure . (1000*) . utcTimeToPOSIXSeconds >>= pure . round

main = do 
    maxns <- getLine 
    let maxn = (read maxns)::Int
    t0 <- getTime 
    loop 1 maxn t0
     where loop n maxn t0|n==maxn = return ()
           loop n maxn t0
             = do 
                 putStrLn $ "fun eval: " ++ (show n) ++ ", " ++ (show $ (f n)) 
                 t <- getTime
                 putStrLn $ "time: " ++ show (t-t0); 
                 loop (n+1) maxn t0