Question

我有一个非常简单的函数f :: Int -> Int，我想为每个f编写一个调用n = 1,2,...,max的程序。每次调用f后，应显示此时使用的（累计）时间（以及n和f n）。如何实施？

我仍然是Haskell中输入/输出的新手，所以这是我到目前为止所尝试的（使用一些玩具示例函数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

我不太清楚如何在这里介绍时间。（Int的{{1}}可能必须替换为其他内容。）

Answer 1

你可能想要这样的东西。根据递归函数的需要调整以下基本示例。

import Data.Time.Clock
import Control.Exception (evaluate)

main :: IO ()
main = do
  putStrLn "Enter a number"
  n <- readLn
  start <- getCurrentTime
  let fact = product [1..n] :: Integer
  evaluate fact  -- this is needed, otherwise laziness would postpone the evaluation
  end <- getCurrentTime
  putStrLn $ "Time elapsed: " ++ show (diffUTCTime end start)
  -- putStrLn $ "The result was " ++ show fact

取消注释最后一行以打印结果（它会很快变得非常大）。

Answer 2

我终于找到了解决方案。在这种情况下，我们用毫秒来衡量“实际”时间。

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

重复计时功能

2 个答案: