Python闭包如何捕获其包含范围的变量？

Question

考虑以下代码：

def foo():
   print('------- entering foo() ---------')

   def internal():
      print('x in internal: {}'.format(x))
      print('id(x) in internal: {}'.format(id(x)))
      print()

   x = 100
   internal()

   global f
   f = internal

   x = 111
   internal()

   print('------- exiting foo() ---------')
   print()


foo()
f()

代码输出为：

------- entering foo() ---------
x in internal: 100
id(x) in internal: 4362053072

x in internal: 111
id(x) in internal: 4362053424

------- exiting foo() ---------
x in internal: 111
id(x) in internal: 4362053424

一旦完成执行，foo()执行的堆栈框架应该被销毁吗？但是，根据上面代码的输出，在我看来，internal引用的f的闭包可以访问分配了{{1 }}。如果栈帧已被破坏，foo()如何访问执行f的栈帧？