foldl is tail recursive, so how come foldr runs faster than foldl?

Question

I wanted to test foldl vs foldr. From what I\'ve seen you should use foldl over foldr when ever you can due to tail reccursion optimization.

This makes sense. Howeve

Answer 1

Well, let me rewrite your functions in a way that difference should be obvious -

a :: a -> [a] -> [a]
a = (:)

b :: [b] -> b -> [b]
b = flip (:)

You see that b is more complex than a. If you want to be precise a needs one reduction step for value to be calculated, but b needs two. That makes the time difference you are measuring, in second example twice as much reductions must be performed.

//edit: But time complexity is the same, so I wouldn't bother about it much.