efficiency question

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