NameError when inputting string instead of int

Question

I am trying to build a simple program that runs the Collatz Sequence on a number inputted by the user. I am also using this to try and get the hang of using 'try except' statements in Python by throwing a print statement whenever the user inputs a string rather than an integer.

The program works as expected when entering a integer, for example here I input the number 12 and I get the following output:

$ python collatzSequence.py 
Enter in a number: 
12
6
3
10
5
16
8
4
2
1

However, when you input a string I get this error:

python collatzSequence.py 
Enter in a number: 
ddd
Traceback (most recent call last):
  File "collatzSequence.py", line 34, in <module>
    main()
  File "collatzSequence.py", line 25, in main
    numb = input()
  File "<string>", line 1, in <module>
NameError: name 'ddd' is not defined

When what I want to be able to do is catch when the user enters something that is not an integer and print 'Please enter an integer.'

The following is my code that I have written, I have placed the 'try except' statement inside the collatz() function, I had also tried putting it in the main() function but I got the same result. :

# Exploring the 'Collatz Sequence'.

# If number is even, print number // 2 and returns this value.
# If number is odd, print and return 3 * number + 1.
def collatz( number ) :
  # Error handling to make sure number entered is an int.
  try:
    number = int(number)
  except NameError :
    print( 'Please enter an integer.')
    return

  # number is even.
  if number % 2 == 0 :
    print( str(number // 2) )
    return number // 2
  # number is odd.
  else :
    print( str(3 * number + 1) )
    return 3 * number + 1

# Lets user type in an integer and keeps calling collatz() on that number until the funciton returns the value 1.
def main() :
  print( "Enter in a number: " )
  numb = input()

  numb = collatz(numb)
  while numb != 1 :
    numb = collatz(numb)



  if __name__ == "__main__":
    main()

Thanks for any help anyone can give me with working out what I can do to fix this issue.

You must put the `try/except` bracket around the line `numb = input()`. It's this line that raises the exception. — DYZ, Jun 05 '18 at 01:50
`input` is evaluating whatever the user enters which is dangerous. `raw_input` just gives you whatever they typed with the newline characters already stripped out. — Ryan Tennill, Jun 05 '18 at 02:17
ah ok, thanks Ryan and Juanpa. I'll make sure to do that instead in the future. — Sly Trash Panda, Jun 05 '18 at 02:21

Sly Trash Panda · Accepted Answer · 2018-06-05T02:20:21.497

Thanks to DyZ for the solution to this. The problem was that I had the 'try except' statement in the wrong spot. I needed to have the 'numb = input()' line in the try statement in the main() function as that is the line that raises the NameError. I also added a while statement so it would continue to ask for an integer if the user keeps inputting a string. This loop breaks once an int is entered. The following is the fixed code :

# Exploring the 'Collatz Sequence'.

# If number is even, print number // 2 and returns this value.
# If number is odd, print and return 3 * number + 1.
def collatz( number ) :
  number = int(number)

  # number is even.
  if number % 2 == 0 :
    print( str(number // 2) )
    return number // 2
  # number is odd.
  else :
    print( str(3 * number + 1) )
    return 3 * number + 1

# Lets user type in an integer and keeps calling collatz() on that number until the funciton returns the value 1.
def main() :
  print( "Enter in a number: " )

  # Error handling to make sure number entered is an int.
  while True :
    try:
      numb = input()
      break
    except NameError :
      print( 'Please enter an integer.' )

  numb = collatz(numb)
  while numb != 1 :
    numb = collatz(numb)

if __name__ == "__main__":
  main()

Giving the result:

$ python collatzSequence.py 
Enter in a number: 
ddd
Please enter an integer.
22
11
34
17
52
26
13
40
20
10
5
16
8
4
2
1

Thanks again DyZ!

NameError when inputting string instead of int

1 Answers1