未绑定变量和名称

未绑定变量和名称

本文介绍了未绑定变量和名称的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

遵循我们拥有的python reference manual

我不知道UnboundLocalError是什么时候引起的?因为

那么我们如何声明一个变量,而不初始化它?

解决方案

您可以引用而不需要为其分配名称:

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

此处引用了foobar,但从未分配.这引起了NameError,因为该名称从未绑定.

更巧妙的是,这里的分配不会发生,因为行永远不会运行:

>>> 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

由于从未执行过spam = 'eggs',因此print spam会引发UnboudLocalError.

请注意,声明在Python中从来没有一个名字.绑定或不绑定,声明不是语言的一部分.

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,import语句,在except子句中保存捕获的异常的名称,在with语句中的上下文管理器的名称所有绑定名称. /p>

如果名称在范围内(例如在函数中)绑定,则它是本地名称,除非您使用global语句(或Python 3中的nonlocal语句)将名称显式标记为而不是全局(或闭包).

以下是错误:

>>> 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foobar功能范围内某处被绑定.但是,如果将foo标记为全局变量,则该函数将起作用:

>>> 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成为全局变量.

所有内容都记录在命名和绑定部分.

Follows to python reference manual we have

I don't to undertand when does UnboundLocalError caused? Because

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

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

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

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

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

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

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.

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.

So the following is an error:

>>> 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

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

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

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

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08-19 20:50