haskell - 缺失的褶皱

Question

如果你想折叠一个列表，我看到了四种方法。

从列表右侧折叠，递归项在右侧

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

从列表右侧折叠，递归项在左侧

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

从列表左侧折叠，递归项在右侧

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

从列表左侧折叠，递归项在左侧

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

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这些折叠中只有两个成为 Prelude asfoldr和foldl。是否有任何理由只包括两个折叠，为什么是这两个？

Answer 1

foldrl并且foldlr不要添加任何表现力：它们与其他两个折叠相同，但折叠功能翻转。

foldrl f = foldr (flip f)
foldlr f = foldl (flip f)

-- Or this, if you prefer
foldrl = foldr . flip
foldlr = foldl . flip

但是用 来定义并不容易foldl，foldr因此同时提供它们是有用的。