How to make the calculator print a statement if the input is a string rather than integer

Question

I am running into a problem, I am trying to make a calculator as my first program!(I am still a very beginner at programming), and i wan't it to print out a statement when someone enters a string rather than an integer , Here is my code:

add = ["add", "+", "plus"]
minus = ["minus", "-", "subtract"]
multiply = ["multiply", "*", "product"]
divide = ["divide", "/",]
quit = ["quit", "q", "leave", "shut down"]
retry = ["retry", "start again", "start", "r",]
leave = False

while True:
    service = input("What would you like to do? ")
    if service.lower() in add:
        x = int(input("First number... "))
        y = int(input("Second number... "))
        print(x + y)
    elif service.lower() in minus:
        x = int(input("First number... "))
        y = int(input("Second number... "))
        print(x - y)
    elif service.lower() in divide:
        x = int(input("First number... "))
        y = int(input("Second number... "))
        print(x / y)
    elif service.lower() in multiply:
        x = int(input("First number... "))
        y = int(input("Second number... "))
        print(x * y)
    elif service != int:
        print("You entered a string , not an integer")
    elif service.lower() in quit:                       #if the user types quit , it'll kill the loop
        break
    else:
        print("Invalid command...Do you want to quit or retry? ")
        lastOption = input()
        if lastOption.lower() in quit:                  #If the user writes in quit , it'll kill the loop here
            break
        elif lastOption.lower() not in retry:           #if the user inputs gibberish again, it begins a check otherwise it restarts
            while True:                                 #A loop begins if the user has entered an unrecognized input
                print("Sorry I did not catch that, can you please repeat? ")
                lastOption = input()
                if lastOption.lower() in quit: 
                    leave = True                        #sets leave condition to true as you cannot break two loops in one break statement
                    break
                elif lastOption.lower() in retry:               
                    break
    if leave == True:                                   #continuing on from the check loop for unrecognized inputs this allows you to break the loops 
        break

EDIT: Here is the "try" and "except" code added

while True:
service = input("What would you like to do? ")
if service.lower() in add:
    try:
        x = int(input("First number... "))
        y = int(input("Second number... "))
        print(x + y)
    except ValueError:
        print("You did not enter an integer")

Welcome to stack overflow! It's always helpful to add example output, with complete errors, when asking a question about error conditions. — erik258, Oct 24 '18 at 16:20

score 0 · Answer 1 · answered Oct 24 '18 at 16:22

0

You're taking input, and casting it to int. The casting will raise a ValueError exception if it isn't an integer, you can try catching that exception.

try:
    x = int(input("First number... "))
    y = int(input("Second number... "))
except ValueError:
    print("You should enter a number instead.")

answered Oct 24 '18 at 16:22

Ahmad Khan

2,655
19
25

Thanks for the help lad, but where do i put this piece of code? – JunkyGardai Oct 24 '18 at 16:34
Wherever you're taking input in `x` and `y`. – Ahmad Khan Oct 24 '18 at 16:35
It does not work, i will put in what i wrote in the edit – JunkyGardai Oct 24 '18 at 16:46

score 0 · Answer 2 · answered Oct 24 '18 at 16:23

0

Write a try / catch block. If a string is passed rather than an integer, it will throw an exception (a flag used to denote an error, halts execution). Wrap the block in a try statement. Then write a catch statement to do something (probably print) when it's not an integer. Of course, there are many types of exceptions, and in this case, if a string is passed, a ValueError exception will be thrown.

try:
    int (input (">> "))
catch ValueError:
    print("Error: string not valid. Try again")

answered Oct 24 '18 at 16:23

Huzaifa Shaikh

24
2

Thanks for the response , but where do i write this block? – JunkyGardai Oct 24 '18 at 16:33
When you're dealing with input, like taking in two operands, put that in the try block, and handle the error in the catch block. – Huzaifa Shaikh Oct 24 '18 at 16:39
IT does not work, i will put in what i wrote in the edit – JunkyGardai Oct 24 '18 at 16:46

How to make the calculator print a statement if the input is a string rather than integer

2 Answers2