Python Square Root Calculator Error

Question

I wanted to make a simple square root calculator.

num = input('Enter a number and hit enter: ')

if len(num) > 0 and num.isdigit():
    new = (num**0.5)
    print(new)
else:
    print('You did not enter a valid number.')

It doesn't seem as if I have done anything wrong, however, when I attempt to run the program and after I have input a number, I am confronted with the following error message:

Traceback (most recent call last):
File "/Users/username/Documents/Coding/squareroot.py", line 4, in <module>
new = (num**0.5)
TypeError: unsupported operand type(s) for ** or pow(): 'str' and 'float'

Process finished with exit code 1

You carefully check if the input could be converted to a number, then don't actually bother to do that! — jonrsharpe
– jonrsharpe, Commented Mar 21, 2017 at 6:36
Also, numbers like 1.5 or 1E10 will not be valid - why? The pythonic way would be to try converting the input to a float and to print the error message only when an exception occurs. — Tim Pietzcker
– Tim Pietzcker, Commented Mar 21, 2017 at 7:12
Possible duplicate of TypeError: unsupported operand type(s) for -: 'str' and 'int' — Łukasz Rogalski
– Łukasz Rogalski, Commented Mar 21, 2017 at 7:16

Tim Pietzcker · Accepted Answer · 2017-03-21 07:29:22Z

3

You can use this solution. Here try and catch is capable of handling all kinds of input. So your program will never fail. And since the input is being converted to float. You won't face any type related error.

try:
    num = float(input('Enter a positive number and hit enter: '))
    if num >= 0:
        new = (num**0.5)
    print(new)

except:
    print('You did not enter a valid number.')

edited Mar 21, 2017 at 7:29

Tim Pietzcker

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answered Mar 21, 2017 at 7:02

Ashish

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Comments

Shivkumar kondi · Accepted Answer · 2017-03-21 06:41:42Z

0

Input function returns you string value. so you need to parse it properly

num = raw_input('Enter a number and hit enter: ')

if num.isdigit():
  if int(num) > 0:
      new = (int(num)**0.5)
      print(new)
else:
    print('You did not enter a valid number.')

edited Mar 21, 2017 at 6:41

answered Mar 21, 2017 at 6:34

Shivkumar kondi

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Klaus D. Over a year ago

Then, his second line will fail.

rocketsnstuff Over a year ago

Hmm, I tried it and it ruled out that error. However, I am now give another error message: Traceback (most recent call last): File "/Users/username/Documents/Coding/squareroot.py", line 3, in <module> if len(num) > 0 and num.isdigit(): TypeError: object of type 'int' has no len()

eyllanesc Over a year ago

@ShivkumarKondi What is the difference between my answer and yours?

Shivkumar kondi Over a year ago

@eyllanesc I just did parsing in first attempt and encounter isdigit problem. You can see the difference. I have executed and then updated my code.

eyllanesc Over a year ago

In the original question, that task is already in place, I do not put it for that. I still see you copied my answer.

Dheeraj · Accepted Answer · 2017-03-21 07:50:39Z

0

Use Math module for simple calculations. refer: Math module Documentation.

import math
num = raw_input('Enter a number and hit enter: ')

if num.isdigit():
    num = float(num)
    new = math.sqrt(num)
    print(new)
else:
    print('You did not enter a valid number.')

answered Mar 21, 2017 at 7:50

Dheeraj

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