Haskell错误约束中的非类型变量参数:Num [E]

因为你foldl可能使用错误的方式.foldl作为签名:

Foldable t => (b -> a -> b) -> b -> t a -> b

所以你给它一个函数(这里(+))一个初始值和一系列值.你可以修改你的代码:

module Asig1 where

f as ys x = (s,z)
       where
       ws = zipWith (*) as ys
       s = foldl (+) x ws --instead of foldl (+) ws x
       z = s

您可以进一步改进代码,因为z = s没有必要:

module Asig1 where

f as ys x = (s,s)
       where
       ws = zipWith (*) as ys
       s = foldl (+) x ws --instead of foldl (+) ws x