Even if the first command in if is true it doesn't print what i want , it only print the else command

Question

numberchk=(int(input(\"Enter a Roman numeral or a Decimal numeral:\" )))
def int2roman(number):
    numerals={1:\"I\", 4:\"IV\", 5:\"V\", 9: \"IX\", 10:\"X\", 40:\"XL\",         


        

Answer 1

if numberchk==int:

This will check if numberchk is equal to the int type. It will not check if numberchk is an integer (which you probably want to do). The correct way to check its type would be this:

if isinstance(numberchk, int):

However, this won’t make sense either. The way you get numberchk is by calling int() on a string:

numberchk=int(input(…))

So numberchk will always be an int. However, calling int() on a string that is not a number can fail, so you probably want to catch that error to find out whether or not the input was a number:

try:
    numberchk = int(input("Enter a Roman numeral or a Decimal numeral:"))
except ValueError:
    print('Entered value was not a number')

But this will again will be problematic, as—at least judging by the message you’re printing—you also want to accept Roman numerals, which can’t be converted to integers by int. So you should also write a function that takes a Roman numeral and converts it into an int.