I’m trying to figure out how to make a percent chance system in Python and can’t figure out why my code won’t work

Question

The code itself is using the variable x which starts as 0 and num which is x % 2.

The code takes x and adds 1 or 2 to x depending on user input. An example of the functionality is:

if num != int: print('hi')

and a contrasting statement:

elif num == int: print("hello")

The code always prints "hi" no matter if num is a whole number or a fraction.

Welcome to Stack Overflow. In your own words, why should `num` ever be equal to `int`? (Hint: is a blue ball *equal to* the colour blue? What would that even mean?) — Karl Knechtel, Feb 02 '22 at 03:21

score 0 · Answer 1 · answered Feb 02 '22 at 03:17

0

You want to use isinstance:

if not isinstance(num, int):

int is a type. You want to see if your number has that type, not if your number is the type int.

answered Feb 02 '22 at 03:17

Kraigolas

Red · Answer 2 · 2022-02-02T03:23:00.800

What you're looking for is the built-in type() function:

num = 5

if type(num) != int: 
    print('hi') 
elif type(num) == int: # Ideally, this should just be "else:"
    print("hello")

Output:

hello

Do note that there exists another built-in function, isinstance(), that can do the same thing:

num = 5

if isinstance(num, int): 
    print('hello') 
else: 
    print('hi')

Output:

hello

BUT, you should keep in mind that this would also return True if num were the boolean value of True:

>>> isinstance(True, int)
True

2 Answers2