error handling - Python excepting input only if in range

Question

Hi I want to get a number from user and only except input within a certain range.

The below appears to work but I am a noob and thought whilst it works there is no doubt a more elegant example... just trying not to fall into bad habits!

One thing I have noticed is when I run the program CTL+C will not break me out of the loop and raises the exception instead.

while True:
  try:
    input = int(raw_input('Pick a number in range 1-10 >>> '))
    # Check if input is in range
    if input in range(1,10):
      break
    else:
      print 'Out of range. Try again'
  except:
    print ("That's not a number")

All help greatly appreciated.

This seems like two completely separate questions... It's best to ask one question at a time. If you get two answers: one to the first question, and another to the second, how do you know which to accept? — Mark Byers
Do you know that range(1,10) doesn't include 10? Your raw_input text suggests that you might want 10 to be acceptable. — DSM
Yes I knew 10 was out of the range, my mistake was to include it in the raw_input statement. Sorry just quickly put values in as an example. Thanks. — hemmy

dav1d dav1d · Accepted Answer · 2012-07-21T18:03:24

Ctrl+C raises a KeyboardInterruptException, your try … except block catches this:

while True:
   try:
       input = int(raw_input('Pick a number in range 1-10 >>> '))
   except ValueError: # just catch the exceptions you know!
       print 'That\'s not a number!'
   else:
       if 1 <= input < 10: # this is faster
           break
       else:
           print 'Out of range. Try again'

Generally, you should just catch the exceptions you expect to happen (so no side effects appear, like your Ctrl+C problem). Also you should keep the try … except block as short as possible.

error handling - Python excepting input only if in range

2 Answers