10 Sep
2010
10 Sep
'10
11:29 p.m.
On Friday 10 September 2010 11:13:50 pm michael rice wrote:
Which of these would be more costly for a long list?
f :: [Int] -> [Int] f [x] = [x] f (x:xs) = x + (head xs) : f xs
f :: [Int] -> [Int]
f [x] = [x] f (x:y:xs) = x + y : f (y:xs)
Another option would be: f [x] = [x] f (x:xs@(y:_)) = (x + y) : f xs However, I believe I've done tests in the past, and your second example generates the same code when optimizations are on (that is, it doesn't build a new y:xs, but reuses the existing one), and that should perform the same as your first implementation. All that said, I'm not sure you'd be able to see the difference anyway. -- Dan