Python递归实现输入数字各位求和时返回None的问题排查

Hey there! Let's break down why your recursive digit sum code is returning None and fix it up. I noticed two critical issues in your implementation:

Key Problems in Your Code #

• Missing return statement in the recursive branch: When inputNum >= 10, you call digit_sum(inputNum / 10) but don't return the result of this call (plus the current digit). Without a return here, the function defaults to returning None once this branch finishes executing. That's exactly why your output shows None for multi-digit numbers.
• Incorrect division type: Using / performs floating-point division (e.g., 123 / 10 gives 12.3), but your function is designed to work with integers. You need integer division (//) to drop the last digit and get a whole number (e.g., 123 // 10 gives 12).

Corrected Code #

Here's the fixed version of your code with clear explanations:

def run():
    inputNum = int(input("Enter an int: "))
    print(f"sum of digits of {inputNum} is {digit_sum(inputNum)}.")

def digit_sum(inputNum):
    # Base case: single-digit number, return the number itself
    if inputNum < 10:
        return inputNum
    # Recursive case: add last digit to sum of remaining digits
    else:
        last_digit = inputNum % 10
        remaining_digits = inputNum // 10
        return last_digit + digit_sum(remaining_digits)

if __name__ == "__main__":
    run()

How It Works #

Let's walk through an example with input 123 to see the recursion in action:

1. digit_sum(123) calculates 3 + digit_sum(12)
2. digit_sum(12) calculates 2 + digit_sum(1)
3. digit_sum(1) hits the base case and returns 1
4. Working back up: 2 + 1 = 3, then 3 + 3 = 6
5. The final result 6 is returned and printed correctly

Now if you run this code, multi-digit inputs will return the correct sum instead of None!

内容的提问来源于stack exchange，提问作者indian_trash