check number that are non-negative integer and not alphabet

Question

i want to continue the loop for asking input(), i use "type(p) is not int" to check the alphabet number, but i get UnboundLocalError when i use "type(p) is not int"

def check(p):
    """
>>> get the value which is non-negative integer and not alphabet
    Checking if the input is negative or not.
    or repeat asking for input
    loop will terminate until positive integer
    """
    while p < 0 or type(p) is not int:
        p = float(input("Invaild response, please try again:"))
    return p

check("4")
check(4)

i want to continue the loop for asking input(), i use "type(p) is not int" to check the alphabet number, but i get UnboundLocalError when i use "type(p) is not int"

Please try to use correct upper case letters, e.g. in the beginning of your title, sentences or the word "I". This would be gentle to your readers. — buhtz, Nov 21 '22 at 14:03
Python doesn't know what `p` is at the while loop. It is only created on the next line. Perhaps you should try adding `p = -1` on the line before the loop. — quamrana, Nov 21 '22 at 14:04
`p` is not defined at the first iteration. Also the check for `type(p) is not int` is a bit useless as `p = float(...)` so it will ***always*** be true... — Tomerikoo, Nov 21 '22 at 14:12
The argument `4` or `"4"` to the function is unused and does nothing…?! — deceze, Nov 21 '22 at 14:14

Jamiu S. · Answer 1 · 2022-11-21T17:20:33.823

def check(num):
    while type(num) is not int or num < 0:
        try:
            num = int(input("Invaild response, please try again:"))
        except ValueError:
            pass
    return num

      
check("4")

Outputs:

print(check(-4))

#Output: Invaild response, please try again:

print(check(4.2))

#Output: Invaild response, please try again:

print(check("4"))

#Output: Invaild response, please try again:

print(check(4))

#Output: 4

check number that are non-negative integer and not alphabet

1 Answers1