Welcome to the Treehouse Community

Want to collaborate on code errors? Have bugs you need feedback on? Looking for an extra set of eyes on your latest project? Get support with fellow developers, designers, and programmers of all backgrounds and skill levels here with the Treehouse Community! While you're at it, check out some resources Treehouse students have shared here.

Looking to learn something new?

Treehouse offers a seven day free trial for new students. Get access to thousands of hours of content and join thousands of Treehouse students and alumni in the community today.

Start your free trial

Python Python Basics (2015) Letter Game App Even or Odd Loop

Ikechukwu Arum
Ikechukwu Arum
5,337 Points
Posted by Ikechukwu Arum
Ikechukwu Arum
5,337 Points

Please help! Error message: wrong number of print.

import random

start = 5

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. #return not num % 2 print("{} is odd".format(num))

while start >5 : num = random.randint(1,99) even_odd(num) start -=1

even.py 
import random

start = 5

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.
    #return not num % 2
    print("{} is odd".format(num))

while start >5 :
    num = random.randint(1,99)
    even_odd(num)
    start -=1

2 Answers

Manish Giri
Manish Giri
16,266 Points
Manish Giri
Manish Giri
16,266 Points

A couple of things -

  1. You have start = 5 at the top, but your while loop condition is while start > 5. Since start is not greater than 5, the while loop will never execute.
  2. It looks like you've modified the even_odd method itself. You were required to use this method in your while loop to check if the random number was odd or even, then print accordingly. With the changed method that you have - 
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.
    #return not num % 2
    print("{} is odd".format(num))

If the number is even, you'll get two prints instead of one, since there's no return after the if block. It prints the statement in the if, and then the one at the last line too.

Ikechukwu Arum
Ikechukwu Arum
5,337 Points
Ikechukwu Arum
Ikechukwu Arum
5,337 Points

thanks for the reply. I understand what you were saying about the while loop condition but i don't understand what you mean by getting two prints and etc

Manish Giri
Manish Giri
16,266 Points

Let's take an example - 

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.
    #return not num % 2
    print("{} is odd".format(num))

even_odd(4)

If you run this code (here, even_odd is exactly the same as what you have), you'll get two print statements - 

4 is even
4 is odd

Why?

When num is even, like in this case with 4, this if block executes- 

 if num %2 ==0: 
        print("{} is even".format(num))

and you get the first print 4 is even.

But then the function continues executing, because there's no return or break. The next line of code encountered after the if block is print("{} is odd".format(num)). Which is why you get 4 is odd, even though 4 is not odd.

To fix this - 

def even_odd(num):
    if num %2 ==0: 
        print("{} is even".format(num))
    else:
        print("{} is odd".format(num))
Ikechukwu Arum
Ikechukwu Arum
5,337 Points
Ikechukwu Arum
Ikechukwu Arum
5,337 Points

ok i figured it out. Thanks for your help