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

Christopher Parke
Christopher Parke
21,978 Points
Posted by Christopher Parke
Christopher Parke
21,978 Points

Even_odd

I'm not sure what I'm missing here...

even.py 
import random

start = 5

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

while start > 4:
    number = random.randint(1, 99)
    if even_odd(number) % 2 != 0:
      print("{} is even".format(number))
      start = start -1
    elif even_odd(number) % 2 = 0:
      print("{} is odd".format(number))
      start = start -1

2 Answers

Steven Parker
Steven Parker
231,248 Points
Steven Parker
Steven Parker
231,248 Points

Your while loop should run until start is "falsey". Yours runs while start > 4 (just one time).

:point_right: You can check for "truthyness" by just naming a variable.

The even_odd function gives you True for even numbers, False for odd ones.

:point_right: You won't need to do any comparisons or math on the result, just test it.

:point_right: Lastly, if you've already tested for even, you can use a plain "else" to handle odd.

Kourosh Raeen
Kourosh Raeen
23,733 Points
Kourosh Raeen
Kourosh Raeen
23,733 Points

A few points:

1) start would be falsey once it's decremented down to zero so just use while start

2) The even_odd() function tells you whether a number is even or odd so you don't need to use %2 in the loop. Just write

if even_odd(number):

3) If it is not even then it is odd so no need for elif. Use else instead.

4) Decrement start just once at the end of the loop