python中赋值操作都会在当前作用域的locals()内部新绑定一个变量。因此如下的代码会报错:

def func1():
    x = 1
    print locals()
    def func2():
        print 'fun2:', locals()
        x += x
        print 'fun2:', locals()

    func2()
    print locals()

if __name__ == '__main__':
    func1()

报错就是

UnboundLocalError: local variable 'x' referenced before ssignment

但是为什么把x改成列表,然后以下面的方式：

def func1():
    x = [1, 2, 3]
    print locals()
    def func2():
        print 'fun2:', locals()
        x[0] += x[0]
        print 'fun2:', locals()

    func2()
    print locals()

if __name__ == '__main__':
    func1()

输出是:

{'x': [1, 2, 3]}
fun2: {'x': [1, 2, 3]}
fun2: {'x': [2, 2, 3]}
{'func3': <function func3 at 0x7f89da00ac08>, 'x': [2, 2, 3], 'func2': <function func2 at 0x7f89da00ab90>}

但是如果把x[0] += x[0]改成x = x[0]又报错了。
求解释啊。。。已经懵了---