将列表中的所有数字相乘 - Haskell

Question

我是haskell的新手,想要将数组中的所有数字相乘.例如.:

阵:

[3,2,4]   //3*2*4

产量

24 

谢谢,非常感谢任何帮助.

Answer 1

在Haskell中有很多方法可以做到这一点.例如,您可以使用:

product [3,2,4]

或者等价的

foldr (*) 1 [3,2,4]

或者递归地:

prod [] = 1
prod (x : xs) = x * prod xs 

函数foldr就是所谓的列表变形.要理解foldr,我们需要首先理解列表构造函数.在Haskell中,[3,2,4]是一个语法糖3 : 2 : 4 : [],其中:是list-cons构造函数,[]是空列表.应用程序foldr f v将按:功能f和空列表替换列表中的每个匹配项v.其定义如下:

foldr :: (a -> b -> b) -> b -> [a] -> b
foldr f v [] = v                       -- equation 1
foldr f v (x:xs) = f x (foldr f v xs)  -- equation 2

例如,考虑foldr (*) 1 [3,2,4]:

foldr (*) 1 [3,2,4]     =
3 * (foldr (*) 1 [2,4]) = (by equation 2 of foldr)
3 * (2 * (foldr (*) 1 [4])) = (by equation 2 of foldr)
3 * (2 * (4 * (foldr (*) 1 []))) = (by equation 2 of foldr)
3 * (2 * (4 * 1)) = (by equation 1 of foldr)
= 24