Python nonlocal statement in a class definition [duplicate]

This question already has answers here: Short description of the scoping rules? (9 answers) The scope of names defined in class block doesn't extend to the methods' blocks. Why is that? (2 answers) Closed 5 months ago.

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 execute开发者_StackOverflow中文版s 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' found

while the following code runs perfectly

def A():
    v = 1
    class B:
        nonlocal v
        v = 2

Can anyone explain the difference here between the closure of the function definition and the class definition?

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Lexical scoping applies only to function namespaces, otherwise methods defined inside a class would be able to "see" the class level attributes (which is by design - those attributes must instead be accessed as attributes of self inside the method).

The same limitations that cause the class level variables to be skipped over by references from methods also keep the nonlocal keyword from working its magic. (global does work though, since that doesn't rely on the lexical scoping machinery)

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Python handles class and function definitions rather differently. For example, your A.v is not a variable of A but rather an attribute of it. The namespace created by a class is not, therefore, a scope. I am not surprised that nonlocal does not work as you're trying to use it.