recursive function to add digits in python fails when leading digit is zero
python, recursion
Solution
In Python 2, integer literals that start with zero are octal.
To take your examples:
In [46]: 012
Out[46]: 10
In [47]: 0123
Out[47]: 83
In [48]: 0010
Out[48]: 8
Since your function works in base ten, it is doing its job correctly. :)
As an aside, you need neither string manipulation nor recursion for this problem. Since others have already suggested non-recursive solutions, here is a recursive one that doesn't use string manipulation:
def sumOfDigits(n):
return 0 if n == 0 else sumOfDigits(n // 10) + n % 10
Problem
I'm trying to create a recursive function that adds all the digits in a number. Here's what I've come up with: ``` def sumOfDigits(num): num=str(num) if len(num)==0: return 0 elif len(num)==1: return int(num) elif len(num)>1: return int(num[0]) + int(num[-1]) + int(sumOfDigits(num[1:-1])) ``` this seems to work for almost any number: ``` sumOfDigits(999999999) >>>81 sumOfDigits(1234) >>>10 sumOfDigits(111) >>>3 sumOfDigits(1) >>>1 sumOfDigits(0) >>>0 ``` strange things happen though if the number begins with '0' ``` sumOfDigits(012) >>>1 sumOfDigits(0123) >>>11 sumOfDigits(00010) >>>8 ``` what am I missing here??