haskell - Finding head of a list using foldl and the last element using foldr

Question

It is known that we can find head of a list using foldr like this:

head'' :: [a] -> a
head'' = foldr (\x _ -> x) undefined

but, is there any way to get the same result using foldl?

Similarly, we can find the last element of list using foldl like this:

last'' :: [a] -> a
last'' = foldl (\_ x -> x) undefined

Is there any way to get the same result using foldr?

标签： haskellfold

Answer 1

head不能用 编写foldl，因为foldl在无限列表上进入无限循环，而head不会。否则，确定：

head' :: [a] -> a
head' = fromJust . foldl (\y x -> y <|> Just x) Nothing

删除fromJust安全版本。

last绝对可以写成 a foldr，大致相同：

last' :: [a] -> a
last' = fromJust . foldr (\x y -> y <|> Just x) Nothing

对于head，我们从 开始Nothing。第一个元素（想要的元素）被包裹Just并用于“覆盖” Nothingwith (<|>)。以下元素被忽略。对于last，它大致相同，但翻转了。