Reputation: 33
I'm trying to make program which counts the number of odd digits in integer using Haskell. I have ran into problem with checking longer integers. My program looks like this at the moment:
oddDigits:: Integer -> Int
x = 0
oddDigits i
| i `elem` [1,3,5,7,9] = x + 1
| otherwise = x + 0
If my integer is for example 22334455 my program should return value 4, because there are 4 odd digits in that integer. How can I check all numbers in that integer? Currently it only checks first digit and returns 1 or 0. I'm still pretty new to haskell.
Upvotes: 0
Views: 607
Reputation: 892
You can first convert the integer 22334455
to a list "22334455"
. Then find all the elements satisfying the requirement.
import Data.List(intersect)
oddDigits = length . (`intersect` "13579") . show
Upvotes: 5
Reputation: 48572
Here's an efficient way to do that:
oddDigits :: Integer -> Int
oddDigits = go 0
where
go :: Int -> Integer -> Int
go s 0 = s
go s n = s `seq` go (s + fromInteger r `mod` 2) q
where (q, r) = n `quotRem` 10
This is tail-recursive, doesn't accumulate thunks, and doesn't build unnecessary lists or other structures that will need to be garbage collected. It also handles negative numbers correctly.
Upvotes: 2
Reputation: 476567
In order to solve such problems, you typically split this up into smaller problems. A typical pipeline would be:
You thus can here implement/use helper functions. For example we can generate a list of digits with:
digits' :: Integral i => i -> [i]
digits' 0 = []
digits' n = r : digits' q
where (q, r) = quotRem n 10
Here the digits will be produced in reverse order, but since that does not influences the number of digits, that is not a problem. I leave the other helper functions as an exercise.
Upvotes: 4