Haskell - how to iterate list elements in reverse order in an elegant way?
haskell, iteration, list, reverse
Solution
The `under reversed f xs` idiom from the `lens` library will apply f to xs in reverse order:
under reversed (take 5) [1..100] => [96,97,98,99,100]
Problem
I'm trying to write a function that given a list of numbers, returns a list where every 2nd number is doubled in value, starting from the last element. So if the list elements are 1..n, n-th is going to be left as-is, (n-1)-th is going to be doubled in value, (n-2)-th is going to be left as-is, etc. So here's how I solved it: ``` MyFunc :: [Integer] -> [Integer] MyFunc xs = reverse (MyFuncHelper (reverse xs)) MyFuncHelper :: [Integer] -> [Integer] MyFuncHelper [] = [] MyFuncHelper (x:[]) = [x] MyFuncHelper (x:y:zs) = [x,y*2] ++ MyFuncHelper zs ``` And it works: ``` MyFunc [1,1,1,1] = [2,1,2,1] MyFunc [1,1,1] = [1,2,1] ``` However, I can't help but think there has to be a simpler solution than reversing the list, processing it and then reversing it again. Could I simply iterate the list backwards? If yes, how?