Having trouble understanding this recursion function

Question

I was reviewing the topic of recursion in a tutorial site and came across the problem and solution below:

Problem:

Given an integer, create a function which returns the sum of all the individual digits in that integer. For example: if n = 4321, return 10 since 4+3+2+1=10.

Solution:

def sum_func(n):
    if n<10:
        return n
    else:
        return n%10 + sum_func(int(n/10))
    pass

I understand the "if n<10" is the base case - if n is single digit, then just return n. What I couldn't wrap my head around is the "n%10 + sum_func(int(n/10))" line of code. Assume n is 235, n%10 will equal 5 and int(n/10) will equal 23. Then ultimately 5 + 23 will become 28 and not 10.

Could someone please help me understand what "n%10 + sum_func(int(n/10))" does? If you could walk through the logic one line at a time that would be greatly appreciated!

int(n/10) is 23, but you are not adding that. Instead, you are adding sum_func(int(n/10)), which is 5 (2+3). — Selcuk
– Selcuk, Commented Nov 6, 2019 at 3:51
n%10 gives last integer 1, 2 3 4 in this order. So try to get digits you need to slice n with int(n/10) this gives 432, 43, 4 so you got 1+2+3+4=10 — Jjj
– Jjj, Commented Nov 6, 2019 at 3:52

Toothpick Anemone · Accepted Answer · 2019-11-06 04:17:29Z

4

if n = 235 then int(n/10) is 23. However, you do not add 5 and 23. You add sum_func(int(n/10)) and 5.

sum_func(int(n/10)) is not int(n/10)

sum_func(int(n/10)) adds the digits in the number "23" together.

As such, sum_func(int(n/10)) == 2 + 3 == 5

sum_func(235) == sum_func(235)
              == sum_func(23)    + 5
              == sum_func(2) + 3 + 5
              == 2           + 3 + 5

answered Nov 6, 2019 at 4:17

Toothpick Anemone

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Stanleyrr Over a year ago

Thanks, Samuel. Your answer was very helpful. I was careless in the beginning and missed the "function within function" and thought I just add 5 and 23 together. But it all makes sense now :)

David Waterworth · Accepted Answer · 2019-11-06 03:59:49Z

2

As you say if there's only 1 digit return it.

The % is the modulus operator - i.e. the remainder when dividing by 10, or simply the last digit (i.e. 235%10 = 5)

int(n/10) drops the last digit since the int() function rounds down - i.e. 235 -> 23

Now what you seem to be confused by is within sum_func it calls sum_func again with the remainder once the last digit has been dropped i.e. 23 (this is the recursion part) which will then return 5.

i.e. you have

sum_func(235)
=5 + sum_func(23)
=5 + (3 + sum_func(2))
=5 + (3 + (2))
=10

answered Nov 6, 2019 at 3:59

David Waterworth

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2 Comments

Stanleyrr Over a year ago

Thank you! I indeed got confused with the sum_func within the sum_func. It's essentially a function within the function. I didn't catch that initially that's why I got confused.

David Waterworth Over a year ago

yeah, that's exactly what recursion is - and if you have to be really careful that one of the branches returns something or you can get infinite recursion (well not really infinite, it'll crash eventually)

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