Python、Javascript中的闭包比较

同为脚本语言，python和Javascript具有相似的变量作用域，不像php，函数的内部的所有变量和外部都是隔绝的，也就是说，函数要想处理其外部的数据，必须使用参数把需要处理的数据传递进来（使用global关键词这里不讨论），而python和Javascript不同，如果在函数声明变量，它会逐级网上查找，直到返回着个值或者未定义。

那么这样说，python的闭包应该很简单了,像javascript一样，我们编写类似的代码：

def func1():
    a = 1
    def func2():
        a = a + 1
        return a
    return func2
re=func1()
print re()
print re()

但是，实际情况是，结果并没有出现我们预期中的打印出2和3,反而出现了这样的错误:”UnboundLocalError: local variable ‘a' referenced before assignment”(局部变量a赋值之前被引用)。为什么会出现这样的问题，我们先看看js是如果实现这个闭包的：

<script>
 function func1(){
 var a=1;
  function func2(){
  a=a+1;
  return a;
  }
 return func2;
 }
re=func1();
console.log(re());
console.log(re());
</script>

上面这段代码的运行结果如我们所料，输入2和3。注意一下这段程序的第5行，如果我在前面加上一个var，这段程序运行的结果是什么样的呢？最终结果是输入了两个“NaN”,在火狐的开发者平台上，找到了关于var这样的描述：

Declares a variable, optionally initializing it to a value.
The scope of a variable declared with var is the enclosing function or, for variables declared outside a function, the global scope (which is bound to the global object).

意思是说，var是用来声明局部变量的，上面的例子中，如果用var a=a+1,这时候的a就已经是func2中的局部变量，而不会从func1中继承，所以最后会出现NaN的结果。

让我们回到python的这个闭包上来，这个错误提示的意思也是说a是个局部变量，实际上，python规定所有在赋值语句左面的变量都是局部变量，这个a在等号左边，所以成了一个局部的变量，导致我访问不到func1中的a。这个问题怎么解决呢？如果是在python3.0以上，在a=a+1之前，可以用nonloacal a来指定a不为局部变量。3.0以下的版本不支持nonloacal关键字,我们可以这样做：

def func1():
    a = [1]
    def func2():
        a[0] = a[0] + 1
        return a[0]
    return func2
re=func1()
print re()
print re()

运行结果如我们所料，打印出了2和3。从python和Javascript闭包的例子，要了解python和js变量声明，变量作用域的相似和不同之处。

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