python - 递归数字求和程序返回 "None"作为 Python 中的总和

Question

这个问题已经有答案了:

Why does my recursive function return None? (4 个回答)

已关闭 11 个月前。

我正在学习递归，并编写了这段代码来返回从 1 到所选数字的所有数字的总和。当我运行程序时，无论我最初输入什么数字，它给我的总和始终是“无”，我的意思是，outputSum变量的值在我的程序中似乎是“无”。我认为更改 SUM_OF_NUMBERS 的初始化值可能会改变这一点，但事实并非如此，我仍然得到“None”作为输出。谁能指出我的代码中导致此错误的部分吗？

def sumOfNumbers(number):

    SUM_OF_NUMBERS = 0

    if number > 0:
        SUM_OF_NUMBERS = SUM_OF_NUMBERS + number
    else:
        return SUM_OF_NUMBERS

    number = number - 1

    sumOfNumbers(number)

def main():

    repeat = 'Y'
    outputSum = 0

    while repeat == 'Y' or repeat == 'y':    
        print("Welcome to the Sum Of Numbers program!")
        number = int(input("\nPlease enter a number to sum up: "))

        outputSum = sumOfNumbers(number)

        print("\nThe sum of all numbers from 1 to " + str(number) + \
              " is " + str(outputSum))

        repeat = input("\nWould you like to sum up another number?" \
                       '\nEnter "Y" for "YES" or "N" for "NO": ')

        if repeat == 'N' or repeat == 'n':
            print("\nThank you for using the program.")
        else:
            print("\nSorry, that was not a valid option.")
            repeat = input('Please enter "Y" for "YES" or "N" for "NO": ')

main()

Answer 1

您得到的是 None，因为当输入大于 0 时您没有返回任何内容。

比你所拥有的简单得多。

def sumOfNumbers(number):
    if number > 0:
        return number + sumOfNumbers(number-1)
    else:
        return 0