python闭合怪异的行为

问题描述

我正在尝试中的问题中的一段代码。

I am trying a piece of code from the question in Lexical closures in Python

flist = []
for i in xrange(3):
    def func(x): return x*i
    flist.append(func)

for f in flist:
    print f.func_closure

输出为：

None
None
None

不应该是？：

(<cell at 0x9222d94: int object at 0x8cabdbc>,)
(<cell at 0x9222d94: int object at 0x8cabdbc>,)
(<cell at 0x9222d94: int object at 0x8cabdbc>,)

我已经使用下面的代码：

I have got the above output using the following code:

flist = []
def actualFact():
    for i in xrange(3):
        def func(x): return x * i
        flist.append(func)

for f in flist:
    print f.func_closure

我使用Python 2.6.6（r266：84292，2010年9月15日，15:52:39）。

I am using Python 2.6.6 (r266:84292, Sep 15 2010, 15:52:39).

推荐答案

只有在全局（模块）范围外引用变量时，才会引入闭包：

Closures are only introduced if there are variables to be referenced outside of the global (module) scope:

>>> def foo():
...     def bar(): pass
...     return bar
...
>>> foo().func_closure is None
True
>>> spam = 'eggs'
>>> def foo():
...     def bar(): return spam
...     return bar
...
>>> foo().func_closure is None
True

只有当内部函数引用变量在周围范围内产生闭包：

Only when the inner function refers to a variable in the surrounding scope are closures generated:

>>> def foo():
...     spam = 'eggs'
...     def bar(): return spam
...     return bar
...
>>> foo().func_closure is None
False
>>> foo().func_closure
(<cell at 0x108472718: str object at 0x108471de0>,)

请注意，您实际上必须引用周围范围中的一个变量。只要忽略范围，您再次无：

Note that you actually have to refer to a variable in the surrounding scope. Simply ignoring the scope gives you None again:

>>> def foo():
...     spam = 'eggs'
...     def bar(): pass
...     return bar
...
>>> foo().func_closure is None
True

在第一个示例中， i 是模块范围变量，只有在第二个示例中，您通过将代码包装在一个新函数中来引入一个新作用域 actualFact 。

In your first example, i is a module-scope variable, only in your second example do you introduce a new scope by wrapping the code in a new function actualFact.

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