当你用foldr在列表中添加数字时,它起作用:
sumIntegers :: [Integer] -> Integer
sumIntegers xs = foldr (+) 0 xs
但是substract的工作方式不同,因为减号=加号。[2,3,4,5]类似于:2-3+4-5。
subtractNums' :: Num a => [a] -> a
subtractNums' xs = foldr (-) 0 xs
subtractNums :: Num a => [a] -> a
subtractNums [] = 0
subtractNums (x:xs) = x - subtractNums xs
有什么需要改变的?提前感谢
加法是结合的(并且是交换的,这意味着恒等式是先加在右边,还是最后加在左边都无关紧要(,所以foldr
和foldl
计算的和相同:
foldr (+) 0 [1,2,3] == 1 + (2 + (3 + 0)) == 1 + 2 + 3 == 6
foldl (+) 0 [1,2,3] == ((0 + 1) + 2) + 3 == 1 + 2 + 3 == 6
然而,减法是而不是关联的;(x - y) - z
通常不等于x - (y - z)
,因此foldr
和foldl
计算两个不同的结果:
foldl (-) 0 [1,2,3] == ((0 - 1) - 2) - 3
== (-1 - 2) - 3
== -3 - 3
== -6
foldr (-) 0 [1,2,3] == 1 - (2 - (3 - 0))
== 1 - (2 - 3)
== 1 - -1
== 2
您选择的函数取决于您实际想要计算的运行差。