even_odd challenge issue

Question

Hi, everyone, I am doing even_odd chanllenge. At first, everything work just fine until the last step. I do everything the question asks for, but it still doesn't work. I know it has to be something with the line 'return not num % 2'. But that also the part I don't understand
If they only ask to pick a random number from 1 to 99 and see whether it is even or odd in 5 turns, what would they need the return statement anyway, because everything is done in even_odd function and the result is printed out straight away, why they need to store the value that 'not num % 2'. Why we need to do that.
In fact, to check whether the 'return not num % 2' has any effect on the code. I test my cod in PyCharm, the code I test doesn't have the return statement and it works beautifully 
import random
def even_odd(num):
    # If % 2 is 0, the number is even.
    if num % 2 == 0:
        print('{} is even.'.format(num))
    # Since 0 is falsey, we have to invert it with not.
    else:
        print('{} is odd.'.format(num))
start = 5
def evenoddgame():
    start = 5
    while start != 0:
        num = random.randint(1, 99)
        even_odd(num)
        start -= 1
evenoddgame()
So can someone explain for me why they include the return statement, and what I can do about it.
Thank you
```even.py
HERE IS WHAT THEY GIVE
def even_odd(num):
    # If % 2 is 0, the number is even.
    # Since 0 is falsey, we have to invert it with not.
    return not num % 2
HERE IS MY ACTUAL CODE
import random
def even_odd(num):
    # If % 2 is 0, the number is even.
    if num % 2 == 0:
        print('{} is even.'.format(num))
    # Since 0 is falsey, we have to invert it with not.
    else:
        print('{} is odd.'.format(num))
        return not num % 2
start = 5
while start != 0:
    num = random.randint(1, 99)
    even_odd(num)
    start -= 1
```

Mike Wagner · Answer

The return statement exists in even_odd to return the result (in this case a True or False) of the operation back to the function call (your even_odd(num)). Essentially, you should leave the even_odd() function alone and write all of your code inside the while loop, though it can be accomplished in any number of other ways. Since the Challenge is expecting it to be done with the even_odd() function left untouched, we'll approach it that way.
By moving your code 
python
if num % 2 == 0:
        print('{} is even.'.format(num))
else:
        print('{} is odd.'.format(num))

down into the while statement below your call to even_odd() and changing the num % 2 == 0 portion to be your function call like so:
```python
start = 5
while start != 0:
    num = random.randint(1, 99)
if even_odd(num):
    print('{} is even'.format(num))
else:
    print('{} is odd'.format(num))

start -= 1
```
You'll achieve the results you need.
*** Edit *** I missed that you had an extra "." in your print strings, so it wasn't passing because of that as well. If you remove them like above, you'll be golden.

Sam Bui · Answer

Oh, I think i get it now, so they mean  that when num % 2 is 0, the evenodd(num) becomes true and it will trigger
if evenodd(num): to print out '{} is even'
if num % 2 not equal 0, then even_odd(num) is false, in this case is 'else:' it will then say '{} is odd'
that makes sense, thank you for your help

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Sam Bui

Sam Bui

even_odd challenge issue

2 Answers

Mike Wagner

Mike Wagner

Sam Bui

Sam Bui

Mike Wagner

Mike Wagner