未绑定变量和名称 [英] Unbound variable and name
问题描述
遵循我们拥有的python reference manual
根本找不到名称时，将引发NameError异常.如果 名称是指尚未绑定的局部变量，a 引发UnboundLocalError异常. UnboundLocalError是一个子类 的NameError.
我不知道UnboundLocalError是什么时候引起的?因为
Python缺少声明，并且允许进行名称绑定操作 在代码块中的任何地方.
那么我们如何声明一个变量，而不初始化它?
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foo在bar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
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.
I don't to undertand when does UnboundLocalError caused? Because
Python lacks declarations and allows name binding operations to occur anywhere within a code block.
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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