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

Ryan Walton
Ryan Walton
1,298 Points

Even and odd Task 3/3 Help!! (Task 1 is no longer working)

I do not know where i am going wrong with the code.

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_number > 0:
    random_num = random.randint(1, 99)
    answer = even_odd(random_num)
    if answer == 0:
        print("{} is even".format(answer))
        start -= 1
        sys.exit()
    else:
        print("{} is odd".format(answer))
        start -= 1
        sys.exit()
Ryan Walton
Ryan Walton
1,298 Points

i fixed up the naming of "start_number" to "start", didn't seem to do the trick

1 Answer

Couple of things...

import random

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

start = 5

# in the condition you probably meant 'start' not start_number
#while start_number > 0:
while start > 0:
    random_num = random.randint(1, 99)
    answer = even_odd(random_num)
    #even_odd is  returning not of 0, so its one, i'd rather suggest to check it with true or false. something like [if answer==True:]
    if answer == 1:
        #its the random number that is to be proved as even or else otherwise
        print("{} is even".format(random_num))
        #start -= 1
        #we shouldn't use exit() function, because it terminates the program, rather just break the loop
        #sys.exit()
        #you dont have to 'break' it either. The loop is suppose to run 5 times
        #break
    else:
        #its the random number that is to be proved as even or else otherwise
        print("{} is odd".format(random_num))
        #start -= 1
        #sys.exit()
        #break
    #if we are doing something in both the conditions we can rather specify it outside the loop itself(although is not the problem, just gives a better look)
    start -= 1