python variable scope in nested functions -

I'm reading this about the decorator.

Step 8 , there is a function defined:

  def external (): x = 1 def internal (): print x # 1 return internal   
 and if we move by:  
  & gt; & Gt; & Gt; Foo = External ()> & gt; & Gt; & Gt; Foo.func_closure # doctest: + ELLIPSIS   
 This X does not print according to the explanation:  
 
 Everything works according to Python's scoping rules- X Our function is a local variable in the outer. When moving the inner Python to the inner point x #, the local variable is detected and it appears in the scope of not getting it, which looks for it in an external form.  
  But the point of view about the life of the variable about things about it? Our variable x function is local to external, which means that it only exists when the function is running externally. We are not able to call without an external call, so according to our experiments according to our Python, we should have any kind of internal and perhaps a runtime error.    
 However, I do not really understand what the meaning of the second paragraph is.  
 I understand that internal () gets the value of x but why is it not printed x?  
  UPDATE :  
 Thanks for the answer, now I understand the reason " Return Inner  "There is only one  pointer  for internal () but it is not executed, this is the reason why internal () does not print X because it  to    
 
 
 You are not calling  internal . You have said  external , which gives  internal , but without calling it. If you want to call  internal , then  foo ()  (since you  external  name  foo  ).  
 The paragraph quoted by you is tangent to this issue. You say that you already understand why  internal  gets the value of  x , which is interpreting paragraphs, in fact, if a local variable is nested Function, and nested function is returned, the value of variable is stored with returned function, even if that variable was not activated until it was active. Typically  x  will end after  external , because  x  is only localized on  external  but  External  returns  internal , which is still required to access  x  if it is required to close  x , therefore It can still be accessed later by  internal .