Luhn function in Haskell

Question

I'm currently working through the book Programming in Haskell (which is absolutely amazing so far), but ran into a problem in exercise 4.8.8.

The task is to implement the Luhn algorithm in Haskell, using a help function luhnDouble :: Int -> Int (which doubles a digit and subtracts 9 if the result is greater than 9) and the mod funcion.

Implementing the luhnDouble function was no problem, but I'm struggling to bring both of them into a function of type Int -> Int -> Int -> Int -> Bool.

I have tried doing it two ways:

luhn :: Int -> Int -> Int -> Int -> Bool
luhn w x y z = if (((luhnDouble w) + x + (luhnDouble y) + z) mod 10) == 0 then True
               else False

I receive a type error.

* Couldn't match expected type `(Integer -> Integer -> Integer)
                                -> Integer -> Integer'
              with actual type `Int'
* The function `(luhnDouble w) + x + (luhnDouble y) + z'
  is applied to two arguments,
  but its type `Int' has none
  In the first argument of `(==)', namely
    `(((luhnDouble w) + x + (luhnDouble y) + z) mod 10)'
  In the expression:
    (((luhnDouble w) + x + (luhnDouble y) + z) mod 10) == 0

But am I not giving the functions 4 Ints as arguments and getting a Bool as a result?

I then tried currying the function and using lambda expressions:

luhn :: Int -> (Int -> (Int -> Bool))
luhn = \w ->(\x ->(\y -> (\z -> ((luhnDouble w) + x + (luhnDouble y) + z mod 10))))

But I'm not sure how to bring in the if expression in here to get a Bool value as a result.

Could anyone help me out, and give me a tip how I could solve this?

And your `if` expression appears invalid. You're missing an "else" case. — Carcigenicate, Nov 27 '16 at 18:37
And you're using `mod` as infix, but don't have it wrapped in backticks, so it's being treated as an argument. That's probably your error. — Carcigenicate, Nov 27 '16 at 18:40

score 3 · Accepted Answer · edited Nov 27 '16 at 20:49

luhn :: Int -> Int -> Int -> Int -> Bool
luhn w x y z = if (((luhnDouble w) + x + (luhnDouble y) + z) mod 10) == 0 then True

You didn't give it an else after if.
You are calling prefix mod, not infix `mod`.

Fix:

luhn :: Int -> Int -> Int -> Int -> Bool
luhn w x y z = if (((luhnDouble w) + x + (luhnDouble y) + z) `mod` 10) == 0
                  then True
                  else False

or

luhn :: Int -> Int -> Int -> Int -> Bool
luhn w x y z = if (mod ((luhnDouble w) + x + (luhnDouble y) + z) 10) == 0
                  then True
                  else False

or, a less redundant version:

luhn :: Int -> Int -> Int -> Int -> Bool
luhn w x y z = (mod ((luhnDouble w) + x + (luhnDouble y) + z) 10) == 0

I added an edit. Sorry if it conflicts with your intent. – Carcigenicate Nov 27 '16 at 18:46 — Carcigenicate, Nov 27 '16 at 18:46

score 1 · Answer 2 · edited Feb 13 '20 at 04:11

1

luhn :: Int -> Int -> Int -> Int -> Bool
luhn w x y z = (luhnDouble w + x + luhnDouble y + z) `mod` 10 == 0

This will work

edited Feb 13 '20 at 04:11

Brian McKenna

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answered Feb 13 '20 at 00:58

Eric Kwarteng

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Luhn function in Haskell

2 Answers2