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
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.
- 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.
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
>>> type(10) is int True >>> type("10") is int False >>>