The first code snippet prints [0, 3] out.
def func():
a = [0]
def swim():
a.append(3)
# a = [1]+a
return a
return swim()
print(func())
The second code snippet raises error "UnboundLocalError: local variable 'a' referenced before assignment"
def func():
a = [0]
def swim():
# a.append(3)
a = [1]+a
return a
return swim()
print(func())
Is a visible/accessible to function swim after all?
It seems this is a commonly asked question as stated in this link. The reason is that variable a inside swim becomes a local variable as soon as there is an assignment to a. It shadows the external a, and local a is not defined before assignment in function swim, so the error rises.
Thanks for all your guys' answers!
You are appending an element in the first code. The id of a is still same.
But at second code, you are re-defining variable a, which is changing the id of that variable. So that you get UnboundLocalError.
When you do such assignment such as a = [1] + a or a += [1] in a scope, the variable would become local to that scope. In your case, that is function swim().
