# Haskell code to verify credit number

I am new to Haskell, and wrote a script to verify credit number. I did some tests, the script worked, but can it be improved further?

isCreditCardNumber :: String -> Bool
isCreditCardNumber number =
0 == creditCardReminder ( creditCardDouble ( blowupCreditCardNumber ( reverse number)))

blowupCreditCardNumber :: String -> [(Int,Char)]
blowupCreditCardNumber creditCardNumber = zip [1..] creditCardNumber

creditCardDouble :: [(Int,Char)] -> [Int]
creditCardDouble [] = []
creditCardDouble ((index,digit):rest)
| even index = (numberDoubleToList ((*2) $digitToInt digit)) ++ creditCardDouble rest | otherwise = digitToInt digit : creditCardDouble rest numberDoubleToList:: Int -> [Int] numberDoubleToList number | number > 9 = map digitToInt (show number) | otherwise = [number] creditCardReminder :: [Int] -> Int creditCardReminder xs = sum xs mod 10  ## 1 Answer You can eta-reduce blowupCreditCardNumber: blowupCreditCardNumber = zip [1..]  Using composition, you can also make isCreditCardNumber pointfree: isCreditCardNumber = (==0) . creditCardReminder . creditCardDouble . blowupCreditCardNumber . reverse  You don't need to optimize numberDoubleToList by special-casing the one-digit case. numberDoubleToList = map digitToInt . show  The explicit recursion in creditCardDouble can be averted by using library functions that specialize in particular recursive patterns: creditCardDouble = concatMap foo where foo (index, digit) | even index = numberDoubleToList$ (*2) $digitToInt digit | otherwise = [digitToInt digit]  foo's name makes foo look like a crutch, and that is good, because it is one. I would inline definitions that are only used once and do not deserve to be in a library: isCreditCardNumber = (==0) . (mod 10) . sum . concatMap foo . zip [1..] . reverse where foo (index, digit) | even index = map digitToInt$ show $(*2)$ digitToInt digit
| otherwise = [digitToInt digit]


digitToInt digit is used in both cases of foo, and so can be factored out:

isCreditCardNumber = (==0) . (mod 10) . sum . concatMap foo . zip [1..] . reverse where
foo (index, digit) = bar index $digitToInt digit bar index | even index = map digitToInt . show . (*2) | otherwise = pure  In fact, we don't need to generate the index and pass it to foo if all we do with it is put it into bar later: isCreditCardNumber = (==0) . (mod 10) . sum . concatMap foo . zip (cycle [pure, map digitToInt . show . (*2)]) . reverse where foo (doubler, digit) = doubler$ digitToInt digit


foo is almost trivial, lets get rid of it entirely:

isCreditCardNumber = (==0) . (mod 10) . sum . concat . zipWith (\$) (cycle [pure, map digitToInt . show . (*2)]) . map digitToInt . reverse


map digitToInt . show is pure on single digits, so we can factor it out of that list and then even factor out the (*):

isCreditCardNumber = (==0) . (mod 10) . sum . concatMap (map digitToInt . show) . zipWith (*) (cycle [1, 2]) . map digitToInt . reverse


Note that if you know the parity of the length of the credit card number, you can get rid of the reverse.

• wow, mate, this is so amazing!!! looks like I have a long way to learn Haskell. – anru Oct 22 '16 at 1:50
• hi mate, looks like "concatMap (digitToInt . show)" has change to concatMap ((\y -> map digitToInt y) . show) – anru Oct 22 '16 at 3:30
• Ah, of course. concatMap (map digitToInt . show), that is. (Or, if you want, map digitToInt . concatMap show, but that might give a little bit of a different intuition about what the code does?) Let me edit in that missing map in the last code block. – Gurkenglas Oct 22 '16 at 4:38