When i run through my calculator, it gives the following results;
Select operation.
1.Add
2.Subtract
3.Multiply
4.Divide
Enter choice(1/2/3/4):3
Enter first number: 1
Enter second number: 5
Invalid! Input
Can anyone explain to me why it is responding with my else if statement, i'v checked the code many times, plus i have copied paste the code directly, as is, after much frustration, yet it yields the same result?
# A simple calculator that can add, subtract, multiply and divide.
# define functions
def add(x, y):
"""This function adds two numbers"""
return x + y
def subtract(x, y):
"""This function subtracts two numbers"""
return x - y
def multiply(x, y):
"""This function multiplies two numbers"""
return x * y
def divide(x, y):
"""This function divides two numbers"""
return x / y
# Take input from the user
print ("Select operation.")
print ("1.Add")
print ("2.Subtract")
print ("3.Multiply")
print ("4.Divide")
choice = input("Enter choice(1/2/3/4):")
num1 = int(input("Enter first number: "))
num2 = int(input("Enter second number: "))
if choice == '1':
print(num,"+",num2,"=", add(num1,num2))
elif choice == '2':
print(num1,"-",num2,"=", subtract(num1,num2))
elif choice == '3':
print(num1,"*",num2,"=", multiply(num1,num2))
elif choice == '4':
print(num1,"/",num2,"=", divide(num1,num2))
else:
print("Invalid! Input")
You're using Python 2, where input() evaluates what is entered; so when you enter 2, for instance, choice contains the int 2. Try entering '2' into your current code (including the quotes). It'll act as you expect entering 2 to act.
You should use raw_input() on Python 2 and input() on Python 3. If you want your code to be compatible with both, you can use the following code, after which you can always just use input():
try:
input = raw_input # Python 2
except NameError: # We're on Python 3
pass # Do nothing
You can also use the six package, which does this and many other Python 2/3 compatibility things.
In Python 3 input() does what raw_input() does in Python 2, and Python 2's input() is gone.
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