Python编程中运用闭包时所需要注意的一些地方
写下这篇博客，起源于Tornado邮件群组的这个问题how to use outer variable in inner method，这里面老外的回答很有参考价值，关键点基本都说到了。我在这里用一些有趣的例子来做些解析，简要的阐述下Python的闭包规则，首先看一个经典的例子:
def foo(): a = 1 def bar(): a = a + 1 # print a + 1 # b = a + 1 # a = 1 print id(a) bar() print a, id(a)
在Python2.x上运行这个函数会报UnboundLocalError: local variable 'a' referenced before assignment即本地变量在引用前未定义，如何来理解这个错误呢？PEP 227里面介绍到，Python解析器在搜索一个变量的定义时是根据如下三级规则来查找的：
The Python 2.0 definition specifies exactly three namespaces to check for each name — the local namespace, the global namespace, and the builtin namespace.
这里的local实际上可能还有多级，上面的代码就是一个例子，下面通过对代码做些简单的修改来一步步理解这里面的规律：
- 如果将a = a + 1这句换成print a + 1或者b = a + 1，是不会有问题的，即在内部函数bar内，外部函数foo里的a实际是可见的，可以引用。
- 将a = a + 1换成 a = 1也是没有问题的，但是如果你将两处出现的a的id打印出来你会发现，其实这两个a不是一回事，在内部函数bar里面，本地的a = 1定义了在bar函数范围内的新的一个局部变量，因为名字和外部函数foo里面的变量a名字相同，导致外部函数foo里的a在内部函数bar里实际已不可见。
- 再来说a = a + 1出错是怎么回事，首先a = xxx这种形式，Python解析器认为要在内部函数bar内创建一个新的局部变量a，同时外部函数foo里的a在bar里已不可见，而解析器对接下来对右边的a + 1的解析就是用本地的变量a加1，而这时左边的a即本地的变量a还没有创建(等右边赋值呢)，因此就这就产生了一个是鸡生蛋还是蛋生鸡的问题，导致了上面说的UnboundLocalError的错误。
要解决这个问题，在Python2.x里主要有两个方案：
用别名替代比如b = a + 1，内部函数bar内只引用外部函数foo里的a。
将foo里的a设成一个容器，如list
def foo(): a = [1, ] def bar(): a[0] = a[0] + 1 bar() print a[0]
当然这有些时候还是很不方便，因此在Python3.x中引入了一个nonloacal的关键字来解决这个问题，只要在a = a + 1前加一句nonloacal a即可，即显式的指定a不是内部函数bar内的本地变量，这样就可以在bar内正常的使用和再赋值外部函数foo内的变量a了。
在搜索Python闭包相关的材料中，我在StackOverflow上发现一个有趣的有关Python闭包的问题，有兴趣的可以思考思考做做看，结果应该是什么？你预期的结果是什么，若不一致，如果要得到你预期的结果应该怎么改？
flist = [] for i in xrange(3): def func(x): return x * i flist.append(func) for f in flist: print f(2)
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