未绑定的变量和名称

2022-01-14 00:00:00 python variables namespaces

问题描述

根据python参考手册我们有

当根本找不到名称时，会引发 NameError 异常.如果该名称指的是尚未绑定的局部变量，a引发 UnboundLocalError 异常.UnboundLocalError 是一个子类的名称错误.

When a name is not found at all, a NameError exception is raised. If the name refers to a local variable that has not been bound, a UnboundLocalError exception is raised. UnboundLocalError is a subclass of NameError.

我不明白什么时候抛出 UnboundLocalError?因为

I don't to understand when the UnboundLocalError is thrown? Because

Python 缺少声明并允许发生名称绑定操作代码块内的任何位置.

Python lacks declarations and allows name binding operations to occur anywhere within a code block.

那么我们怎样才能声明一个变量，而不是初始化她呢?

So how we can declare a variable, but not to initialize her?

解决方案

你可以引用一个没有分配的名字:

You can refer to a name without having assigned to it:

>>> foobar
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
NameError: name 'foobar' is not defined

这里 foobar 被引用，但从未被分配到.这会引发 NameError 因为名称从未绑定.

Here foobar is being referred to, but was never assigned to. This raises a NameError because the name was never bound.

更微妙的是，这里没有发生分配，因为执行的行永远不会运行:

More subtly, here assignment is not happening because the line that does is never run:

>>> def foo():
...     if False:
...         spam = 'eggs'
...     print spam
... 
>>> foo()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 4, in foo
UnboundLocalError: local variable 'spam' referenced before assignment

因为 spam = 'eggs' 永远不会执行，所以 print spam 会引发 UnboudLocalError.

Because spam = 'eggs' is never executed, print spam raises an UnboudLocalError.

请注意，Python 中没有任何地方声明过.您绑定或不绑定，声明不是语言的一部分.

Note that nowhere in Python is a name ever declared. You bind or don't bind, declaration is not part of the language.

相反，binding 用于确定名称的范围；绑定操作包括赋值、用于 for 循环的名称、函数参数、导入语句、在 except 子句中保存捕获的异常的名称、上下文管理器的名称一个 with 语句所有的绑定名称.

Instead, binding is used to determine the scope of a name; binding operations include assignment, names used for a for loop, function parameters, import statements, name to hold a caught exception in an except clause, the name for a context manager in a with statement all bind names.

如果名称绑定在范围内(例如在函数中)，则它是本地名称，除非您使用 global 语句(或 nonlocal 语句在 Python 3 中)将名称显式标记为全局(或闭包).

If a name is bound in a scope (such as in a function) then it is a local name, unless you use a global statement (or a nonlocal statement in Python 3) to explicitly mark the name as a global (or a closure) instead.

所以下面是一个错误:

>>> foo = None
>>> def bar():
...     if False:
...         foo = 'spam'
...     print foo
... 
>>> bar()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 4, in bar
UnboundLocalError: local variable 'foo' referenced before assignment

因为 foo 被 somewhere 绑定在 bar 函数范围内.但是如果您将 foo 标记为全局，则该函数可以工作:

because foo is being bound somewhere in the bar function scope. But if you mark foo as a global, the function works:

>>> foo = None
>>> def bar():
...     global foo
...     if False:
...         foo = 'spam'
...     print foo
... 
>>> bar()
None

因为现在 Python 编译器知道您希望 foo 改为全局变量.

because now the Python compiler knows you wanted foo to be a global instead.

所有这些都记录在 命名和绑定部分 Python 参考文档.

This is all documented in the Naming and Binding section of the Python reference documentation.

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