Write a recursive Python function, given a non-negative integer N, to calculate and return the sum of its digits

Question

I've tried this but it doesn't work.

def sumDigits(n):
    if n == 0:
        print 0
    else:
        print n%10 + sumDigits(n/10)
sumDigits(35)

One of the key parts of recursion is returning values from the recursive function. You might want to check [this answer](http://stackoverflow.com/a/30214677/1903116) to know more about the basics of recursion. — thefourtheye, Jul 10 '15 at 14:43
Are you sure that you want to *print* the answer (as opposed to returning it)? — John Coleman, Jul 10 '15 at 14:48
Have you considered adding `return`? This isn't a tutorial service. — jonrsharpe, Jul 10 '15 at 14:56
Try formatting according to PEP 8; e.g. function names should be all lowercase. — noɥʇʎԀʎzɐɹƆ, Jul 10 '15 at 14:57

score 2 · Accepted Answer · answered Jul 10 '15 at 14:50

If you are using Python 3.x, note that you need to use // to perform integer division, as / will perform float (true) division even with two integers.

In any case, the function needs to return the values, not just print them.

def sumDigits(n):
    if n == 0:
        return 0
    else:
        return n%10 + sumDigits(n//10)

Then it will work as intended

>>> sumDigits(35)
8

Wayne Treible · Answer 2 · 2015-07-10T14:59:20.143

Here is a solution that works for numbers (and strings!) of all lengths.

def sumDigits(n):
    n = str(n)

    if n is "":
        # Base Case: Return 0 when there is nothing left
        return 0    
    else:
        # Default Case: Return the first number + sum of the rest)
        return int(n[0]) + sumDigits(n[1:])    


>>> print sumDigits(359)
>>> 17

This, when written out, will look like this:

sumDigits(359) = 3 + sumDigits(59) = 3 + 5 + sumDigits(9) = 3 + 5 + 9 + 0

Rahul Gupta · Answer 3 · 2015-07-10T15:08:44.440

You need to return values instead of printing values.

Try this:

def sumDigits(n):
    if n == 0:
        return 0
    else:
        return n%10 + sumDigits(n//10)

>>> sumDigits(41)
5

>>> sumDigits(31)
4

You can also use and operator to leverage the equality check when n is 0.

def sumDigits(n):
    return n and n%10 + sumDigits(n//10)

Here, n%10 + sumDigits(n//10) will be evaluated only when n is non-zero.

Use // as it gives the quotient on division in which the digits after the decimal point are removed.

score 0 · Answer 4 · answered Jul 10 '15 at 15:01

0

When you attempt to make the recursive call sumDigits( n / 10) it expects to have a value returned. When it fails to get that value because the function doesn't return anything, you receive the error.

This could prove useful: Python Functions

answered Jul 10 '15 at 15:01

NamesJ

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Write a recursive Python function, given a non-negative integer N, to calculate and return the sum of its digits

4 Answers4