how to print out something instead of geting an error when list.pop function is out of range?

My training course task is to make a shoppingcart, everything works as intented but I can’t figure out the function pop out of range problem.

My code:

# -*- coding: cp1252 -*-
class Cart:

    shoppingcart = []

    def addstuff(self):
        esine = input("What will be added?: ")
        self.shoppingcart.append(esine)

    def remove(self):
        esine2 = input("Which item is deleted?: ")
        self.shoppingcart.pop(int(esine2))

def main():
    customer = Cart()
    while True:
        selection = input("Would you like to \n"
                          "(1)Add or \n"
                          "(2)Remove items or \n"
                          "(3)Quit: ")

        if selection == "1":
            customer.addstuff()

        if selection == "2":
            print("There are", len(customer.shoppingcart), "items in the list.")
            customer.remove()

        if selection == "3":
            print("The following items remain in the list:")
            for i in \
                    customer.shoppingcart:
                print(i)

            break

        if selection > "3":
            print("Incorrect selection.")


if __name__ == "__main__":
    main()

Problem is while deleting(poping) an item that is out of range it gives an error but instead should print out "incorrect selection."
I tried adding if True and else argument into def remove part but that doesn’t help.

>Solution :

Use try except blocks.

emptylist = []
try:
    emptylist.pop()
except IndexError:
    print('pop failed')

In your code:

def remove(self):
    esine2 = input("Which item is deleted?: ")
try:
    self.shoppingcart.pop(int(esine2))
except IndexError:
    print('no item at index {}'.format(int(esine2))

Do note that pop(i) removes the item in the list at place i, not the item with value i.