[Haskell-beginners] any feedback on this solution
Kees Bleijenberg
K.Bleijenberg at lijbrandt.nl
Sat Feb 7 09:56:11 UTC 2015
Roelof,
Maybe it's better to reuse toDigits in sumDigits.
Kees
-----Oorspronkelijk bericht-----
Van: Beginners [mailto:beginners-bounces at haskell.org] Namens Roelof Wobben
Verzonden: zaterdag 7 februari 2015 10:37
Aan: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell
Onderwerp: [Haskell-beginners] any feedback on this solution
Hello,
I finally solved exercise 1 where I had to write a programm which checks
if a creditcard number is valid.
I solved it this way :
toDigits :: Integer -> [Integer]
toDigits n
| n < 0 = []
| n < 10 = [n]
| otherwise = toDigits (n `div` 10) ++ [n `mod` 10]
-- | convert a number to a reversed array where a negative number will
be a empty array
toDigitsRev :: Integer -> [Integer]
toDigitsRev 0 = [0]
toDigitsRev n
| n < 0 = []
| n < 10 = [n]
| otherwise = n `mod` 10 : toDigitsRev (n `div` 10)
-- | Doubles every second number from the right.
doubleEveryOther :: [Integer] -> [Integer]
doubleEveryOther [] = []
doubleEveryOther (x:[]) = [x]
doubleEveryOther (x:(y:zs))
| length (x:(y:zs)) `mod` 2 /= 0 = [x] ++ (y * 2) : doubleEveryOther zs
| otherwise = [x *2] ++ y : doubleEveryOther zs
-- | sum all the digits of a array
sumDigits :: [Integer] -> Integer
sumDigits [] = 0
sumDigits (x:zs)
| x < 10 = x + sumDigits zs
| otherwise = x `mod` 10 + x `div` 10 + sumDigits zs
-- | validate a number by looking if a number can be divided by 10
validate :: Integer -> Bool
validate n = sumDigits(doubleEveryOther(toDigits(n))) `mod` 10 == 0
-- | The main entry point.
main :: IO ()
main = do
print $ validate 4012888888881881
Any remarks about this solution.
I know there are higher function solutions but that part is not
descrived in chapter 1 .
Roelof
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