Question

我的代码应该从n倒数到1.代码完成但最后返回None。有关为什么会发生这种情况以及如何解决问题的任何建议？提前致谢！

def countdown(n):

    '''prints values from n to 1, one per line
    pre: n is an integer > 0
    post: prints values from n to 1, one per line'''

    # Base case is if n is <= 0
    if n > 0:
        print(n)
        countdown(n-1)
    else:
        return 0

def main():

    # Main function to test countdown
    n = eval(input("Enter n: "))
    print(countdown(n))

if __name__ == '__main__':
    main()

Answer 1

print(countdown(n))打印countdown(n)返回的值。如果n > 0，返回的值为None，因为如果没有执行return语句，Python函数默认返回None。

由于您要在countdown内打印值，因此修复代码的最简单方法是删除print中的main()来电：

def countdown(n):

    '''prints values from n to 1, one per line
    pre: n is an integer > 0
    post: prints values from n to 1, one per line'''

    # Base case is if n is <= 0
    if n > 0:
        print(n)
        countdown(n-1)

def main():

    # Main function to test countdown
    n = eval(input("Enter n: "))
    countdown(n)

if __name__ == '__main__':
    main()

编辑：删除else-clause，因为docstring表示最后打印的值是1，而不是0。

Answer 2

删除print(countdown(n))并将其替换为countdown(n)。

当n大于0时，倒计时函数会返回任何内容，因此print语句正在打印None。您也可以在countdown(n)中删除您的其他声明 - 因为您从不关心倒计时的返回值，所以它没有用处。

def countdown(n):

    '''prints values from n to 1, one per line
    pre: n is an integer > 0
    post: prints values from n to 1, one per line'''

    # Base case is if n is <= 0
    if n > 0:
        print(n)
        countdown(n-1)
    # No need to return a value from here ever - unused in the recursion and
    # you don't mention wanting to have zero printed.
    #else:
    #    return 0

def main():

    # Main function to test countdown
    n = eval(input("Enter n: "))

    # Don't print the result of countdown - just call it
    #print(countdown(n))
    countdown(n)

if __name__ == '__main__':
    main()

Answer 3

您希望将递归调用返回到countdown的下一次调用：

def countdown(n):

    '''prints values from n to 1, one per line
    pre: n is an integer > 0
    post: prints values from n to 1, one per line'''

    # Base case is if n is <= 0
    if n > 0:
        print(n)
        return countdown(n-1)
    else:
        return 0

def main():

    # Main function to test countdown
    n = eval(input("Enter n: "))
    print(countdown(n))

if __name__ == '__main__':
    main()

这样，当达到基本情况时，返回值0应该向上传播回堆栈并返回。

不幸的是，因为你的基本情况返回0，这导致打印0.但是，如果你想要返回一些东西（想想你是否真的需要在这种情况下）这将是你如何做它

如果您不需要返回任何内容，则无需打印结果。进一步修改

print(countdown(n))

到

countdown(n)

Python递归函数在完成后返回none

3 个答案: