将每个函数应用于列表Haskell的每个元素

Question

I have a list of a function like this:

[(+1), (+2), (*4), (^2)]

And i want to apply each function to each element of an another list. For example i have a list like this [1..5], i want to get this as result: [2,4,12,16]

This is what i tryed already.

applyEach :: [(a -> b)] -> [a] -> [b]
applyEach _ [] = []
applyEach (x:xs) (y:ys) = x y : applyEach xs ys

I don't know what is the problem, we have a online surface where we have to place the code and it test our submision, and only said that my code isn't pass.

Answer 1

Your function works fine when the lists are the same length, or when the second list is shorter than the first:

> applyEach [(+1), (+2), (*4), (^2)] [1..4]
[2,4,12,16]
> applyEach [(+1), (+2), (*4), (^2)] [1..3]
[2,4,12]

But you're not dealing with the case where the second list is longer, as it is in your example:

> applyEach [(+1), (+2), (*4), (^2)] [1..5]
[2,4,12,16*** Exception: H.hs:(2,1)-(3,47): Non-exhaustive patterns in function applyEach

You need to add one more equation to your function to handle this case.

Answer 2

您也可以使用内置的zipWith函数和$运算符来执行此操作：

applyEach fs xs = zipWith ($) fs xs