The local names for a function are decided when the function is defined:
>>> x = 1
>>> def inc():
... x += 5
...
>>> inc.__code__.co_varnames
('x',)
In this case, x
exists in the local namespace. Execution of x += 5
requires a pre-existing value for x
(for integers, it's like x = x + 5
), and this fails at function call time because the local name is unbound - which is precisely why the exception UnboundLocalError
is named as such.
Compare the other version, where x
is not a local variable, so it can be resolved at the global scope instead:
>>> def incg():
... print(x)
...
>>> incg.__code__.co_varnames
()
Similar question in faq: http://docs.python.org/faq/programming.html#why-am-i-getting-an-unboundlocalerror-when-the-variable-has-a-value