運行python doc示例無法理解python作用域和名稱空間一章中的內容？

Question

我正在學習python3.6文檔，當我查看python scopes and namespaces ，我嘗試運行此代碼，我發現在scope_test()調用do_local()打印結果與我的想法不同：

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)

    do_nonlocal()
    print("After nonlocal assignment:", spam)

    do_global()
    print("After global assignment:", spam)

scope_test()
print("In global scope:", spam)

我認為在調用do_local()時應在do_local()范圍內查找垃圾郵件，因為do_local()范圍具有spam變量，因此：

do_local()
print('After local assignment:', spam) # local spam

除非do_local()沒有spam變量，然后可以在scope_test()找到spam

但是python解釋器的打印結果是：

do_local()
print('After local assignment:', spam) # test spam

do_local

與調用do_global()時類似，我認為在此范圍內，因為global spam是全局的，所以：

do_global()
print("After global assignment:", spam) # test spam

但是為什么結果是：

do_global()
print("After global assignment:", spam) # nonlocal spam

Answer 1

def scope_test():
    def do_local():
        spam = "local spam" #this spam variable is in do_local's scope
    def do_nonlocal():
        nonlocal spam #this spam variable refers to spam variable defined in outer function's ,that is scope_test's,scope
        spam = "nonlocal spam"
    def do_global():
        global spam #this refers spam variable defined outside all functions (global)
        spam = "global spam"

    spam = "test spam" #this spam variable is defined in scope_test's scope

    do_local()
    print("After local assignment:", spam) #this spam var is still in scope_test's scope

    do_nonlocal() #do_nonlocal is called and inside that spam variable (defined nonlocal) is changed which changes scope_test's spam
    print("After nonlocal assignment:", spam) #this print is also in scop_test's scope but above function changed current scope's value 

    do_global() #changes spam var in global scope 
    print("After global assignment:", spam) #prints spam var in scop_test

scope_test() #execution starts
print("In global scope:", spam) #prints spam var in global scope as this print statement is in global scope

Answer 2

這是我運行程序后得到的結果。 讓我們經歷一下。

def do_local():
    spam = "local spam"

將創建對象垃圾郵件，該垃圾郵件的范圍僅限於此函數do_local，並且在函數壽命到期時將其銷毀。 因此，在本地分配后，您會看到測試垃圾郵件的輸出，該輸出是使用spam = "test spam"定義的，該spam = "test spam"的范圍與函數scope_test有關。

def do_nonlocal():
    nonlocal spam
    spam = "nonlocal spam"

通過使用nonlocal關鍵字，您將告訴解釋器您正在使用的對象不限於功能do_nonlocal的范圍。 因此，解釋器在此函數范圍之外搜索該對象，並在函數scope_test找到它。 然后，它將與功能scope_test的范圍相關的“非本地垃圾郵件”的值分配給該垃圾郵件對象。

def do_global():
    global spam
    spam = "global spam"

在這里，您實際上是在創建全局對象垃圾郵件，並為其分配“全局垃圾郵件”的值。 請記住：這是整個程序范圍的全局范圍。 但是，當您打印出垃圾郵件時，python會查找垃圾郵件對象並僅在函數范圍內找到它，例如，在do_nonlocal函數調用中分配了“非本地垃圾郵件”值的垃圾郵件，因此您將非本地垃圾郵件視為輸出。

scope_test()
print("In global scope:", spam)

運行范圍測試后，最初包含“測試垃圾郵件”值的垃圾郵件對象的范圍在函數scope_test結束后仍然存在。 因此，當您嘗試再次打印出垃圾郵件時，全局對象垃圾郵件將被打印出來，該垃圾郵件是您在do_global中調用do_global時test_scope 。

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我希望它能清除您的困惑！