It say UnboundLocalError, How can i update variable after assigning it in this code?

Question

name = "Aayan"

def AI():
    print("Hi, {}. I hope you are doing well :)".format(name))
    ask = input("How can i help you, {} ? : ".format(name))
    if ask == "change name" or "rename":
        new_name = input("What should I call you ? : ")
        name = new_name
    else:
        exit()

AI()

It throws UnboundLocalError: local variable 'name' referenced before assignment. How can I update name after asking it from the user ?

You should return the value from the function with `return name` and call the function with `name = AI()` so you rebind `name` in the outer scope. Don't use `global`. Never use exit. To cite the documentation of [`exit`](https://docs.python.org/3/library/constants.html?highlight=exit#exit) (and `quit`): "They are useful for the interactive interpreter shell and should not be used in programs." — Matthias, Jun 11 '21 at 12:45
Does this answer your question? [UnboundLocalError on local variable when reassigned after first use](https://stackoverflow.com/questions/370357/unboundlocalerror-on-local-variable-when-reassigned-after-first-use) — joanis, Jun 11 '21 at 13:02

Rishabh Semwal · Accepted Answer · 2021-06-11T12:50:40.847

0

name = "Aayan"

def AI():
    global name
    print("Hi, {}. I hope you are doing well :)".format(name))
    ask = input("How can i help you, {} ? : ".format(name))
    if ask == "change name" or "rename":
        new_name = input("What should I call you ? : ")
        name = new_name
        return name
    else:
        exit()


print(AI())

use global keyword inside function and call name

edited Jun 11 '21 at 12:50

answered Jun 11 '21 at 12:37

Rishabh Semwal

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Now it throws SyntaxError: name 'name' is used prior to global declaration. – AaYan Yasin Jun 11 '21 at 12:41
@AaYanYasin Now check the code. Tell me if it not work – Rishabh Semwal Jun 11 '21 at 12:51

It say UnboundLocalError, How can i update variable after assigning it in this code?

1 Answers1