Python: modify nonlocal variable after return from outer function

Question

Consider the following example:

def g():

    in_g=100

    def f1():
        nonlocal in_g
        in_g = 10

    def f2():
        nonlocal in_g
        print(in_g)

    return (f1,f2)


(f1, f2) = g()

f2() #prints 100
f1()
f2() #prints 10

Both the inner functions, f1 and f2, have, in their "closures", access to the variable in_g. However, after g returns, where is in_g kept in the memory?

I assume that while g is executing, in_g is a variable on the stack frame corresponding to a call to g. Therefore, g, f1 and f2 all access the same memory location (on the stack) when using the variable in_g.

But, as seen in the example, after g returns, f1 and f2 still access the same memory location when referencing in_g. However, now that g returned, that memory location cannot be on the stack anymore.

Answer 1

I believe I found the answer here: http://stupidpythonideas.blogspot.ro/2015/12/how-lookup-works.html.

So, when accessing in_g, g, f1 and f2 access a cell variable which, in turn, holds a reference to the actual object.