[Haskell-cafe] types and number of evaluation steps

[Victor Gorokgov]

+ on Int is extremely cheap. It is always faster to add again rather 
than store the value.
But Integer is a different story. Addition time on this type can grow to 
several minutes.

18.02.2012 13:28, Heinrich Hördegen пишет:
>
> Dear all,
>
> I have a question about evaluation with respect to types and currying. 
> Consider this programm:
>
> import Debug.Trace
>
> -- add :: Integer -> Integer -> Integer
> add :: Int -> Int -> Int
> add x y = x + y
>
> f a b c = trace "b" (add x c) where x = trace "a" (add a b)
>
> main :: IO ()
> main = do
> print (f 1 2 3)
> print (f 1 2 4)
>
>
> Compiled with ghc-7.0.3:
>
> $ ghc --make Main.hs -o main -O2
>
> The function add has to types. When we use type Int -> Int -> Int, the 
> programm produces "b a 6 b a 7" as output which shows that the x from 
> the where clause in f is evaluated twice. However, when we use type 
> Integer -> Integer -> Integer, this will give "b a 6 b 7" which shows 
> that x is evaluated only once. This was rather unexpected to me.
>
> Why does the number of evaluation steps depend on a type? Can anybody 
> explain this or give a hint?
>
> Thank you very much,
> Heinrich
>
>
>