How to pass a list received in one function to another function - in Haskell?

I'm writing a function that checks if the mod of a (transformed) list is equal to zero or not, I'm using two functions, but I don't know how I can pass the list received in the "valida" function to the "duplicador" function.

To work, I'm passing the list inside the "valida" function, but not as a parameter.

I wish that where the duplicador [1, 3, 8, 6] I could pass the argument/list received in the "valida" function

module Main where


duplicador :: [Integer] -> [Integer]
duplicador [] = []
duplicador (x:[]) = [x]
duplicador (x:y:zs)
  |  (length (x:y:zs)) `mod` 2 /= 0 =  x : y*2 : duplicador zs
  |  otherwise = x*2 : y : duplicador zs




valida :: [Integer] -> Bool
valida r
    | r == 0 = True
    | otherwise = False
        where
        r = foldr (\x y -> x y) 0 (duplicador [1, 3, 8, 6]) `mod` 10
        

main = do

 print(valida [1, 3, 8, 6])```

CodePudding user response：

You are shadowing the argument r in valida. Renaming r to something let's you use it.

valida :: [Integer] -> Bool
valida entrada
    | r == 0 = True
    | otherwise = False
        where
        r = foldr (\x y -> x y) 0 (duplicador entrada) `mod` 10

We can also improve the code a bit further. foldr (\x y -> x y) 0 can be simplified to foldr ( ) 0, but still it is a very common pattern, we can replace it with sum. And for the pattern if bool then true else false is simply bool, rendering:

valida r = sum (duplicador r) `mod` 10 == 0