Encountering a NameError related to a free variable can be quite daunting. This is a scarier version of the <a href="https://pythonprohub.com/python-errors/unboundlocalerror-local-variable-referenced-before-assignment/">UnboundLocalError</a> we saw earlier. It happens specifically when dealing with nested functions (a function inside another function).
The Scenario
You have an outer function with a variable, and an inner function tries to change it.
Problem Code:
def outer():
x = 10
def inner():
# We try to change 'x' from the outer scope
x += 1
print(x)
inner()
outer()
# CRASH! NameError: free variable 'x' referenced before assignment...The Fix: nonlocal
Just like we used global to fix scope issues at the top level, we use nonlocal to fix them in nested functions. It tells Python: “I don’t want a new local variable; I want the one from the outer (non-global) scope.”
def outer():
x = 10
def inner():
nonlocal x # <--- The magic fix
x += 1
print(x)
inner()
outer()
# Output: 11 (Success!)
