simple calculator using python

Question

Hello I am new to python.i am just going through basic python tutorial and making a simple calculator in python but its not working.here it is:

#calculator program
#this variable tells the loop whether it should loop or not.
# 1 means loop. anything else means don't loop.

loop = 1

#this variable holds the user's choice in the menu:

choice = 0

while loop == 1:
#print what options you have
print ("Welcome to calculator.py")

print ("your options are:")
print (" ")
print ("1) Addition")
print ("2) Subtraction")

print ("3) Multiplication")

print ("4) Division")
print ("5) Quit calculator.py")
print (" ")

choice = input("Choose your option: ")
if choice == 1:
    add1 = input("Add this: ")
    add2 = input("to this: ")
    print (add1, "+", add2, "=", add1 + add2)
elif choice == 2:
    sub2 = input("Subtract this: ")
    sub1 = input("from this: ")
    print (sub1, "-", sub2, "=", sub1 - sub2)
elif choice == 3:
    mul1 = input("Multiply this: ")
    mul2 = input("with this: ")
    print (mul1, "*", mul2, "=", mul1 * mul2)
elif choice == 4:
    div1 = input("Divide this: ")
    div2 = input("by this: ")
    print (div1, "/", div2, "=", div1 / div2)
elif choice == 5:
    loop = 0

print ("Thankyou for using calculator.py!")

by pressing F5 to run its coming

Welcome to calculator.py
your options are:

1) Addition
2) Subtraction
3) Multiplication
4) Division
5) Quit calculator.py

Choose your option:

by choosing 1 it should prompt for input number but its going back and showing

Welcome to calculator.py
your options are:

i think it cannot intercept my option choise when entered 1 or 2 or 3 or 4 or 5 and going back.what is wrong with it

In 3.x `input` *always returns a string*. `'1' != 1`. I suggest you implement [a sub-function to ensure valid input](http://stackoverflow.com/questions/23294658/asking-the-user-for-input-until-they-give-a-valid-response). — jonrsharpe, Jul 10 '14 at 15:18

score 1 · Accepted Answer · answered Jul 10 '14 at 15:18

1

choice = input("Choose your option: ")

when you enter choice as 1. It takes 1 as string but, your comparing string with int. Simply cast your input to int

choice = int(input("Choose your option: "))

answered Jul 10 '14 at 15:18

Ashoka Lella

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ok so by putting int(input("choose option: ") it should work? – Jul 10 '14 at 15:19
you need to use `int(input)` for all input values – Padraic Cunningham Jul 10 '14 at 15:21

simple calculator using python

1 Answers1