python Calculator: Checking input validity

Question

my code seems to work fine except it is not printing the except Exception: print("This input is invalid.") part. When I try to fix it things gets worse. Can any one help me what I am missing? Thank you!

Sample correct output:

Calculator
Give a number: hah, NEVER
This input is invalid.
Give a number: What?
This input is invalid.
Give a number: 100
Give a number: Just kidding
This input is invalid.
Give a number: 50
(1) +
(2) -
(3) *
(4) /
(5)sin(number1/number2)
(6)cos(number1/number2)
(7)Change numbers
(8)Quit
Current numbers: 100 50
Please select something (1-6): 2
The result is: 50
(1) +
(2) -
(3) *
(4) /
(5)sin(number1/number2)
(6)cos(number1/number2)
(7)Change numbers
(8)Quit
Current numbers: 100 50
Please select something (1-6): 8
Thank you!

my code:

def getnumber():
 while True:
 try:
 number = input("Give a number: ")
 if number.isdigit():
 return number
 except Exception:
 print("This input is invalid.")
def main():
 import math
 print("Calculator")
 promptForNumbers = True
 while True:
 if promptForNumbers:
 number1 = int(getnumber())
 number2 = int(getnumber())
 promptForNumbers = False
 print("(1) +\n\n(2) -\n\n(3) *\n\n(4) /\n\n(5)sin(number1/number2)\n\n(6)cos(number1/number2)\n\n(7)Change numbers\n\n(8)Quit\n")
 print("Current numbers: %s %s" % (number1, number2))
 selection = int(input("Please select something (1-8): ")) 
 if selection == 1:
 print("The result is: %s" % (number1 + number2))
 print("\n")
 elif selection == 2:
 print("The result is: %s" % (number1-number2))
 print("\n")
 elif selection==3:
 print("The result is: %s" % (number1*number2))
 print("\n")
 elif selection==4:
 print("The result is: %s" % (number1/number2))
 print("\n")
 elif selection==5:
 print("The result is: %s" % math.sin(number1/number2))
 print("\n")
 elif selection==6:
 print("The result is: %s" % math.cos(number1/number2))
 print("\n")
 elif selection==8:
 print("Thank you!")
 break
 elif selection==7:
 promptForNumbers = True
 else:
 print("Selection was not correct.") 
if __name__ == "__main__":
 main()

• Matt Ball: What piece(s) of code do you expect to throw an exception?
• user3358884: I want the exception to print "This input is invalid." when input is not a number. That is what I am trying to do. But it just keeps asking for input with out printing "This input is invalid."

Answer 1

The reason why you are not seeing your exception statement executed, is that no code within your try block will raise an exception. The constructor to int will however, raise a ValueError exception if it is passed something that isn't a number.

Based on my understanding, attempting to convert, then catching the exception is the most pythonic way to validate input in this situation. You can read more about exceptions in Python in the docs.

def getnumber():
 while True:
 try:
 return int(input("Give a number: "))
 except ValueError:
 print("This input is invalid.")

You'll want to be sure to remove the calls to int from around all calls to getnumber within main. Also, one last note. It is considered proper form to have your import statements unintended at the top of the file. If you want to learn more you can read about it at PEP8.

Answer 2

Try this:

def getnumber():
 while True:
 number = raw_input("Give a number: ")
 if number.isdigit():
 return number
 else:
 print("This input is invalid.")

Answer 3

 if number.isdigit():
 return number
 else:
 raise ValueError('Not a valid input value.')

• Deepak Ingole: Please explain behavior of change