ValueError exception not working in python

Question

I'm trying to make a simple program that will calculate the area of a circle when I input the radius. When I input a number it works, but when I input something else I'd like it to say "That's not a number" and let me try again instead of giving me an error.

I can't figure out why this is not working.

from math import pi

def get_area(r):
    area = pi * (r**2)
    print "A= %d" % area

def is_number(number):
    try:
        float(number)
        return True
    except ValueError:
        return False

loop = True
while loop == True:
    radius = input("Enter circle radius:")
    if is_number(radius) == True:
        get_area(radius)
        loop = False
    else:
        print "That's not a number!"

Moses Koledoye · Accepted Answer · 2016-11-12 18:06:02Z

1

When you don't input a number, the error is thrown by input itself which is not in the scope of your try/except. You can simply discard the is_number function altogether which is quite redundant and put the print statement in the except block:

try:
    radius = input("Enter circle radius:")
except (ValueError, NameError):
    print "That's not a number!"
get_area(radius)

edited Nov 12, 2016 at 18:06

answered Nov 12, 2016 at 17:59

Moses Koledoye

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Comments

Francisco · Accepted Answer · 2016-11-12 17:59:53Z

0

radius is still a string,

replace

 get_area(radius)

with

get_area(float(radius))

You also have to replace input with raw_input since you're using Python 2

answered Nov 12, 2016 at 17:59

Francisco

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Comments

Feras · Accepted Answer · 2016-11-12 18:04:05Z

0

in= 0
while True:
  try:
     in= int(input("Enter something: "))       
  except ValueError:
     print("Not an integer!")
     continue
  else:
     print("Yes an integer!")
     break

answered Nov 12, 2016 at 18:04

Feras

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