A=[]
def main():
global A
A=[1,2,3,4,5]
b()
def b():
if(len(A)>0):
A=[7,8,9]
else:
if(A[3]==4):
A.remove(2)
main()
This code gives error in line A.remove(2) giving reason:"UnboundLocalError: local variable 'A' referenced before assignment"
but A list is global and for sure it has been initialized in main() function.Please explain why this is giving error?
As A has been initialized again in b function, will this cause error?
The reason you are getting this is because when you performed this assignment in your function:
A = [7, 8, 9]
The interpreter will now see A as a locally bound variable. So, what will happen now, looking at this condition:
if(len(A)>0):
Will actually throw your first UnboundLocalError exception, because, due to the fact, as mentioned, you never declared global A in your b function and you also created a locally bound variable A = [7, 8, 9], you are trying to use a local A before you ever declared it.
You actually face the same issue when you try to do this:
A.remove(2)
To solve this problem with respect to your code, simply declare the global A back in your b() function.
def b():
global A
if(len(A)>0):
A=[7,8,9]
else:
if(A[3]==4):
A.remove(2)
However, the better, ideal and recommended way to do this is to not use global and just pass the arguments to your function
A = [1, 2, 3, 4]
def main(list_of_things):
# do whatever operations you want to do here
b(list_of_things)
def b(list_of_things):
# do things with list_of_things
main(A)
You must declare a variable global in any function that assigns to it.
def b():
global A
if some_condition:
A.append(6)
else:
A.remove(2)
Without declaring A global within the scope of b(), Python assumes that A belongs in the b() namespace. You can read it, but cannot edit it (any changes made will not persist to the true A.
You're allowed to set A to something inside the function A = ... will work, but only within the scope of the function.
If you try to mutate A inside of the function without defining A in your function's namespace, then Python will tell you that you are trying to mutate a variable outside of your purview with UnboundLocalError
By declaring A global, the interpreter knows that you mean to edit the global variable.
I'm assuming that you meant for your code to look something like this (Otherwise, it runs fine)
A=[]
if __name__ == '__main__':
def main():
global A
A=[1,2,3,4,5]
b()
def b():
global A
if some_condition:
A=[7,8,9]
else:
A.remove(2)
print A
main()
print A
print A
