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

Dezhi Zhu
Dezhi Zhu
2,662 Points
Posted by Dezhi Zhu
Dezhi Zhu
2,662 Points

Why is wrong

Make a while loop that runs until start is falsey. Inside the loop, use random.randint(1, 99) to get a random number between 1 and 99. If that random number is even (use even_odd to find out), print "{} is even", putting the random number in the hole. Otherwise, print "{} is odd", again using the random n

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 = True:
    num = random.randint(1,99)
    if even_odd(num) == True:
        print("{} is even.".format(num))

    else:
        print("{} is odd."format(num))

        start= start - 1

1 Answer

Steven Parker
Steven Parker
229,788 Points
Steven Parker
Steven Parker
229,788 Points

Since it is a number, "start" will never be equal to "True". It will, however, be "truthy" while it contains a non-zero value. To test a value for "truthy" you simply name it: "while start:"

Also, the indentation of the line that decrements "start" causes it to be part of the "else" block. But it should be done regardless of the condition.

And while this won't cause a problem, you don't ever need to test a boolean value against "True". Just like testing "truthiness", you can test a boolean value just by naming it.

Dezhi Zhu
Dezhi Zhu
2,662 Points
Dezhi Zhu
Dezhi Zhu
2,662 Points

Thanks million Steven. You enlighten me