[Haskell-cafe] types and number of evaluation steps
Heinrich Hördegen
hoerdegen at funktional.info
Sat Feb 18 10:28:04 CET 2012
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
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www.funktional.info
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