What are the problems encountered in the example of the official python3.6 document scope and namespaces?

def scope_test():
    def do_local():
        spam = "local spam"
    def do_nonlocal():
        nonlocal spam
        spam = "nonlocal spam"
    def do_global():
        global spam
        spam = "global spam"
    spam = "test spam"
    do_local()
    print("After local assignment:", spam) -sharp test spam
    do_nonlocal()
    print("After nonlocal assignment:", spam) -sharp nonlocal spam
    do_global()
    print("After global assignment:", spam) -sharp nonlocal spam

scope_test()
print("In global scope:", spam)
  1. Why do_local () prints test spam instead of local spam
    2. When
  2. executes do_global () , since the variable spam is redeclared as a global variable in this function, why print nonlocal spam instead of test spam .
Python3.x
Mar.28,2021

first, print ("After local assignment:", spam) is always looking for the spam, of the current scope, that is, 

def scope_test():
    ...
    spam = "test spam"  -sharp  spam :spam_4
    ...
    print("After local assignment:", spam) -sharp  spam_4 
    print("After nonlocal assignment:", spam) -sharp  spam_4 
    print("After global assignment:", spam) -sharp  spam_4

answer the first question:
do_local () internal spam scope is only inside this do_local () , so no value is assigned to spam_4, so print test_spam .
answer the second question:
in executing do_nonlocal () , spam_4 has been assigned nonlocal spam . The assignment in
and do_global () does not change the value of spam_4, so print nonlocal spam instead.

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