I tried to make a variable as "private" in the module using __dunder_case__ and use it in the function below.
I finally changed the variable name to uppercase naming and forget about anything private when using python. I'm running python 3.12.2 on windows 11. But I cannot wrap my mind around why is it treated this way...
__use_this__ = 0
def test_out_scope():
__use_this__ += 1
return __use_this__
if __name__ == "__main__":
print(__use_this__)
print(test_out_scope())
print(test_out_scope())
print(test_out_scope())
print(test_out_scope())
and expecting to add the variable but got this instead:
UnboundLocalError: cannot access local variable '__use_this__' where it is not associated with a value.
Edited
__use_this__ = [0]
def test_out_scope():
__use_this__[0] += 1
return __use_this__[0]
if __name__ == "__main__":
print(__use_this__[0])
print(test_out_scope())
print(test_out_scope())
print(test_out_scope())
print(test_out_scope())
expecting to fail
This has nothing to do with variable naming. Double underscore only makes variable names behave differently in class definitions - it is treated like any other variable otherwise.
The reason you're seeing an error when __use_this__ == 0 is simply because of incorrect reference to nonlocal variable. Because integers are immutable - += doesn't work in place with immutable variables. It creates a new object (well, not really but for simplicity let's say it does) and assigns it to the variable name. Since you are reassigning name inside a function, it is treated as local and throws UnboundLocalError: cannot access local variable '__use_this__' where it is not associated with a value.
When you are modifying a list, it still points to the same list object. Since you don't assign it anywhere in the function, it looks for the reference in outer scope.
Simply marking a variable as global will solve the issue.
def test_out_scope():
global __use_this__
__use_this__ += 1
return __use_this__