creating a python program which ask the user for a number - even odd

Question

I am new to python and I am taking a summer online class to learn python.

Unfortunately, our professor doesn't really do much. We had an assignment that wanted us to make a program (python) which asks the user for a number and it determines whether that number is even or odd. The program needs to keep asking the user for the input until the user hit zero. Well, I actually turned in a code that doesn't work and I got a 100% on my assignment. Needless to say our professor is lazy and really doesn't help much. For my own knowledge I want to know the correct way to do this!!! Here is what I have/had. I am so embarrassed because I know if probably very easy!

counter = 1
num = 1
while num != 0:
counter = counter + 1
   num=int(input("Enter number:"))
   while num % 2 == 0:
   print ("Even", num)
else:
    print ("Odd", num)

pls make sure to accurately represent your indentation

patrick
– patrick

2017-07-25 00:18:02 +00:00
Commented Jul 25, 2017 at 0:18 — patrick
– patrick, Commented Jul 25, 2017 at 0:18

Chris · Accepted Answer · 2017-07-25 00:23:49Z

2

There are a couple of problems with your code:

You never use counter, even though you defined it.
You have an unnecessary nested while loop. If the number the user inputs is even, then you're code will continue to run the while loop forever.
You are incorrectly using the else clause that is available with while loops. See this post for more details.

Here is the corrected code:

while True:
    # Get a number from the user.
    number = int(input('enter a number: '))

    # If the number is zero, then break from the while loop
    # so the program can end.
    if number == 0:
        break

    # Test if the number given is even. If so, let the
    # user know the number was even.
    if number % 2 == 0:
        print('The number', number, 'is even')
    # Otherwise, we know the number is odd. Let the user know this.
    else:
        print('The number', number, 'is odd')

Note that I opted above to use an infinite loop, test if the user input is zero inside of the loop, and then break, rather than testing for this condition in the loop head. In my opinion this is cleaner, but both are functionally equivalent.

edited Jul 25, 2017 at 0:23

answered Jul 25, 2017 at 0:18

Chris

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2 Comments

JBarela Over a year ago

Thank you for explaining what I did wrong. I really appreciate your help! Thank You, again

Chris Over a year ago

Glad I could help out @JBarela :-)

Prune · Accepted Answer · 2017-07-25 00:17:46Z

1

You already have the part that continues until the user quits with a 0 entry. Inside that loop, all you need is a simple if:

while num != 0:
    num=int(input("Enter number:"))
    if num % 2 == 0:
        print ("Even", num)
    else:
        print ("Odd", num)

I left out the counter increment; I'm not sure why that's in the program, since you never use it.

answered Jul 25, 2017 at 0:17

Prune

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Comments

Abhishta Gatya · Accepted Answer · 2017-07-25 00:18:00Z

0

Use input() and If its only number specific input you can use int(input()) or use an If/else statement to check

answered Jul 25, 2017 at 0:18

Abhishta Gatya

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Comments

JRG · Accepted Answer · 2017-07-25 00:19:51Z

0

Your code wasn't indented and you need to use if condition with else and not while

counter = 1
num = 1
while num != 0:
    counter = counter + 1
    num = int(input("Enter number:"))
    if num % 2 == 0:
       print ("Even", num)
    else:
       print ("Odd", num)

Sample Run

Enter number:1
Odd 1
Enter number:2
Even 2
Enter number:3
Odd 3
Enter number:4
Even 4
Enter number:5
Odd 5
Enter number:6
Even 6
Enter number:0
Even 0

answered Jul 25, 2017 at 0:19

JRG

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