define test1 l = let s1 = foldr (+) 1 l s2 = foldr (-) 1 l in (s1, s2) test2 l = let s = foldr (\x (a,b) -> (x+a,x-b)) (1,1) l in s why is test1 so much faster than test2 for long lists l (eg [1..1000000])? replacing foldr with foldl makes it faster (of course), but test2 is still much slower. i *expected* test2 to be much faster because you're only traversing the list once. presumably the two elements "a" and "b" in test2 could be put in registers and i'd imagine test2 should be faster (it certainly would be if written in c). - hal -- Hal Daume III "Computer science is no more about computers | hdaume@isi.edu than astronomy is about telescopes." -Dijkstra | www.isi.edu/~hdaume
On Friday 08 February 2002 22:14, Hal Daume III wrote:
define
test1 l = let s1 = foldr (+) 1 l s2 = foldr (-) 1 l in (s1, s2)
test2 l = let s = foldr (\x (a,b) -> (x+a,x-b)) (1,1) l in s
why is test1 so much faster than test2 for long lists l (eg [1..1000000])? replacing foldr with foldl makes it faster (of course), but test2 is still much slower.
i *expected* test2 to be much faster because you're only traversing the list once. presumably the two elements "a" and "b" in test2 could be put in registers and i'd imagine test2 should be faster (it certainly would be if written in c).
- hal
I'd say that's because in the second case you also got to apply the (,), besides the (+)/(-) constructor during the transversing... Am I right? (if you had no mentioned it though, I'd probably also expect the 2nd one to be faster...) J.A.
participants (2)
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Hal Daume III -
Jorge Adriano