def is_it_integer(prompt):
n = prompt
try:
int(n) == n
prompt = True
return n
except ValueError:
print("error.")
prompt = False
return prompt
count = 0
first_guess = input("Enter a number..:")
if is_it_integer(first_guess):
first_guess = int(first_guess)
if count < first_guess:
print("GO TO BED IT WORKED!")
Input these 3 characters before your code (```), left to 1 on my keyboard, but don’t include the brackets Hope this helps
1 Like
tried it, no luck. But thank you.
nevermind, totally worked, was using ‘’’, instead of ```
Nice, also you can do this: [code] … [/code] around the code.
2 Likes
Two things:
- int(n) == n is how you convert an int, however this line doesn’t really do anything. You would need to assign the result to something. x = int(n) == n
- You can really just do: return int(n) and ditch everything inside the try: before the except: If you attempt to convert something not an int it will throw the except, so then in your except: just return False.
1 Like
That looks so much better, and I get it now. Thanks!!!
I’m ignorant here because something I thought would work, doesn’t. I was wondering why we couldn’t just use the type() function to do something like this
if type(x) is int(1):
print('x is an int')
but it fails, with either == or is..
>>> x = 1
>>> type(x) == int(1)
False
>>> type(x)
<class 'int'>
>>> type(int(1))
<class 'int'>
>>> type(x) is int(1)
False
why?
Try this:
>>> type(10) is int
True
>>> type("10") is int
False
>>>