python - Python 中奇怪的多个返回值

Question

通过 Code Academy 学习 Python 仅 5 天。我对任何其他语言一无所知(对 Ruby 知之甚少!)。

这段代码我做错了什么？

<小时/>

Q: Write a function, by_three, that calls a second function, cube, if a number is evenly divisible by 3 and "False" otherwise. You should then return the result you get from cube. As for cube, that function should return the cube of the number passed from by_three. (Cubing a number is the same as raising it to the third power).

So, for example, by_three should take 9, determine it's evenly divisible by 3, and pass it to cube, who returns 729 (the result of 9**3). If by_three gets 4, however, it should return False and leave it at that.

Lastly, call by_three on 11, 12, and 13 on three separate lines.

答:

def by_three(n):
    orig_num = n
    if (isinstance(orig_num, int) and orig_num%3 == 0 ):
        cube(orig_num)
    else:
        print "False"

def cube(orig_num):
    cube = orig_num**3
    print cube
    return

by_three(11)
by_three(12)
by_three(13)

当我运行上面的代码时，这就是我得到的结果。为什么这些值会以这种方式出现？

False
1728
False
==> None
False
False
1728
Oops, try again.

Answer 1

我无法说出为什么你会看到奇怪的结果。当我将您的代码复制到解释器中时，我看到:

>>> def by_three(n):
...     orig_num = n
...     if (isinstance(orig_num, int) and orig_num%3 == 0 ):
...         cube(orig_num)
...     else:
...         print "False"
... 
>>> def cube(orig_num):
...     cube = orig_num**3
...     print cube
...     return
... 
>>> by_three(11)
False
>>> by_three(12)
1728
>>> by_three(13)
False

不过，我认为这个问题比你提出的要简单得多。很难说，因为问题写得很差，但这就是我的答案:

def by_three(n): return False if n % 3 else cube(n)

def cube(n): return n**3

by_three(11)
by_three(12)
by_three(13)

这就是解释器中的样子:

>>> def by_three(n): return False if n % 3 else cube(n)
... 
>>> def cube(n): return n**3
... 
>>> by_three(11)
False
>>> by_three(12)
1728
>>> by_three(13)
False