-1

I've got the following code:

fn = input("Choose a function(1, 2, 3, 4, 5, other(quit)): ");
while (not(fn > '5' or fn < '1')):
    print("hello world");

This works, most of the time. For example, if I input 54 or some crazy number, it will never print "hello world".
However, when I enter 45, it does enter the loop.
Why is this?

Phantômaxx
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Human Cyborg Relations
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  • Because they are compared lexicographically... `'4'` is lexicographically smaller than `'5'` and greater than `'1'`, thus the expression evaluates to `False`. `not False` is `True` thus it goes into the loop. – Andrew Li Oct 29 '16 at 23:24
  • In other words - when using the `>` or ` – Alex Oct 29 '16 at 23:25

3 Answers3

0

Python doesn't end statements with semi-colons

Neither while or not do not require parenthesis.

And you need to compare numbers, not strings. But wait, input returns a string in Python3, so you need to int() that as well. 

fn = input("Choose a function(1, 2, 3, 4, 5, other(quit)): ")
if fn == 'quit':
    # break out of this choose a function prompt
else: 
    while int(fn) in range(1, 5 + 1): # +1 because non-inclusive ranges
       print("hello world")

And, this will be an infinite while loop for a valid input, so just be prepared for that

OneCricketeer
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-1

You're using strings instead of numbers ('5', '1'). String comparison is done by the single character, and the character '4' in '45' counts as less than the character '5'. 

>>> ord('4')
52
>>> ord('5')
53
Roberto
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-1

You could fix it by writing:

fn = input("Choose a function(1, 2, 3, 4, 5, other(quit)): ");
while str(fn) + ',' in '1,2,3,4,5,':
    print("hello world");
Progg
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