python - 关于python中非本地和全局名称的混淆

Question

所以我一直在读一些python toturial，偶然发现了一些我不太明白的东西。
当我将以下代码写入解释器时:

    >>> def scope_test():
    ...    def try_add():
    ...       nonlocal spam
    ...       spam = "new spam"
    ...    try:
    ...       print(spam)
    ...    except:
    ...       print("exception caught")
    ...    try_add()
    ...    print(spam)

我收到以下错误:

      File "<stdin>", line 3
    SyntaxError: no binding for nonlocal 'spam' found

但是当我运行这段代码时:

>>> def scope_test():
...    def try_add():
...       global spam
...       spam = "new spam"
...    try:
...       print(spam)
...    except:
...       print("exception caught")
...    try_add()
...
>>> scope_test()
exception caught
>>> spam
'new spam'

它运行良好。
对此有何解释？

Answer 1

来自文档:https://docs.python.org/3/reference/simple_stmts.html#the-nonlocal-statement

Names listed in a nonlocal statement, unlike those listed in a global statement, must refer to pre-existing bindings in an enclosing scope (the scope in which a new binding should be created cannot be determined unambiguously).

(强调我的)。强调后的陈述告诉你为什么应该这样。
与 global如果变量不存在，那么它可以在全局范围内创建。所以，在你的例子中 global python 愉快地创建 spam在全局范围内。
与 nonlocal , 如果变量不存在，应该在哪个范围内创建？最近的封闭范围和全局范围之间可能有许多范围。而不是猜测，nonlocal坚持变量已经存在。在您的示例中使用 nonlocal ，没有 spam封闭范围和全局范围之间的任何地方，所以你会得到错误。