If you want to fold a list, I see four ways to do it.
Fold from the right of the list, with the recursive term on the right
foldrr (-) 100 [1..10] = 1 - (2 - (3 - (4 - (5 - (6 - (7 - (8 - (9 - (10 - (100)))))))))) = 95
foldrr :: (a -> b -> b) -> b -> [a] -> b
foldrr step zero (x:xs) = step x (foldrr step zero xs)
foldrr _ zero [] = zero
Fold from the right of the list, with the recursive term on the left
foldrl (-) 100 [1..10] = ((((((((((100) - 10) - 9) - 8) - 7) - 6) - 5) - 4) - 3) - 2) - 1 = 45
foldrl :: (a -> b -> a) -> a -> [b] -> a
foldrl step zero (x:xs) = step (foldrl step zero xs) x
foldrl _ zero [] = zero
Fold from the left of the list with the recursive term on the right
foldlr (-) 100 [1..10] = 10 - (9 - (8 - (7 - (6 - (5 - (4 - (3 - (2 - (1 - (100)))))))))) = 105
foldlr :: (a -> b -> b) -> b -> [a] -> b
foldlr step zero (x:xs) = foldlr step (step x zero) xs
foldlr _ zero [] = zero
Fold from the left of the list with the recursive term on the left
foldll (-) 100 [1..10] = ((((((((((100) - 1) - 2) - 3) - 4) - 5) - 6) - 7) - 8) - 9) - 10 = 45
foldll :: (a -> b -> a) -> a -> [b] -> a
foldll step zero (x:xs) = foldll step (step zero x) xs
foldll _ zero [] = zero
Only two of these folds made it into Prelude as foldr and foldl. Was there any reason to just include two folds, and why those two?