haskell - 理解 Haskell 中的折叠

Question

根据我对 Haskell 中折叠的理解，foldl (-) 0 [1..5] 通过计算 0- 给出了 -15 的结果1-2-3-4-5，而foldr (-) 0 [1..5]通过计算得到-5的结果>5-4-3-2-1-0。为什么 foldl (++) ""["a", "b", "c"] 和 foldr (++) ""["a", "b", "c"] 给出 "abc" 的结果，而 foldr 的结果不是 "cba"?

在理解 foldl 和 foldr 之间的区别时，我是否遗漏了什么？

Answer 1

我认为this part from the docs让它更清楚:

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In the case of lists, foldr, when applied to a binary operator, a starting value (typically the right-identity of the operator), and a list, reduces the list using the binary operator, from right to left:

foldr f z [x1, x2, ..., xn] == x1 `f` (x2 `f` ... (xn `f` z)...)

. . .

In the case of lists, foldl, when applied to a binary operator, a starting value (typically the left-identity of the operator), and a list, reduces the list using the binary operator, from left to right:

foldl f z [x1, x2, ..., xn] == (...((z `f` x1) `f` x2) `f`...) `f` xn

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如果您查看示例分解，串联 foldr 等同于:

"a" ++ ("b" ++ ("c" ++ ""))

对于 foldl，它等同于:

(("" ++ "a") ++ "b") ++ "c"

对于字符串连接，这些是相同的。

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但是对于减法，

1 - (2 - (3 - 0))

给出的结果不同于:

((0 - 1) - 2) - 3