[Haskell-beginners] Time interval calculation

[legajid]

After a few efforts, i've understood my mistake.
I now write :
   t1d <- gettime
   putStrLn ("Methode 1 : " ++ show t1d)
   let t1=last (listeprem 5000)
   putStrLn ("            " ++ show t1)
   t1f <- gettime
   putStrLn ("            " ++ show t1f)

and that's ok.

What i feel shocking is having io code mixed within calculation code.
I think it's possible to force calculation of t1d, t1 and t1f  before 
(or without) displaying  their values (i just want t1f - t1d).
But how to make it ?

Thanks,
Didier


.

legajid a écrit :
> Hi,
> my following code should show time before executing listeprem, then 
> time after execution.
>
> import System.Time
> gettime :: IO ClockTime
> gettime = getClockTime
>
> main=do
>    t1d <- gettime
>    let t1=last (listeprem 5000)
>    t1f <- gettime
>    putStrLn ("Methode 1 : " ++ show t1d)
>    putStrLn ("            " ++ show t1)
>    putStrLn ("            " ++ show t1f)
>
> Looking at the screen, t1d is displayed then, after a few seconds, t1 
> and t1f.
> But, t1d and t1f are equal.
> It seems like if t1d and t1f where calculated at start of procedure, 
> before we need calculating t1 for putStrLn.
> How can i have t1f evaluated after t1, so i can calculate time elapsed 
> for calculation of t1?
>
> Thanks,
> Didier
>
>
>
>
>
> _______________________________________________
> Beginners mailing list
> Beginners at haskell.org
> http://www.haskell.org/mailman/listinfo/beginners
>