python-nonlocal

This question already has an answer here: In Python, variables inside if conditions hide the global scope even if they are not executed? 4 answers Today I'm reading python change log and meet the nonlocal keyword and did some experiment with it. I find a confusing situation where untriggered assignment will change the nonlocal keyword behavior, please see the example below. def a(): x = 'a' def b(): def c(): nonlocal x x = 'c' c() b() print(x) a() >>> python3 test.py c def a(): x = 'a' def b(): def c(): nonlocal x x = 'c' c() if False: x = 'b' b() print(x) a() >>> python3 test2.py a You can

I'm trying to perform some analysis of scope in Python 3 source code and I'm stuck with how the nonlocal statement statement works inside a class definition. As I understand it, the class definition executes its body inside a new namespace (call it dict) and binds the class name to the result of type(name, bases, dict). Nonlocal x should work as long as it refers to a variable that is bound somewhere in the enclosing non-local scope. From this I expect the following code to compile and run: class A: v = 1 class B: nonlocal v v = 2 but this fails with SyntaxError: no binding for nonlocal 'v'

I'm trying to implement a closure in Python 2.6 and I need to access a nonlocal variable but it seems like this keyword is not available in python 2.x. How should one access nonlocal variables in closures in these versions of python? Chris B. Inner functions can read nonlocal variables in 2.x, just not rebind them. This is annoying, but you can work around it. Just create a dictionary, and store your data as elements therein. Inner functions are not prohibited from mutating the objects that nonlocal variables refer to. To use the example from Wikipedia: def outer(): d = {'y' : 0} def inner():