python中如何让构造函数变成private的?

[guanliyuanjia]

我希望让构造函数成为private，别人只能通过getMyClass()来获取实例。
请问怎么做？
谢谢！

[xyzxyz1111]

试用decorator

import traceback
def private(func):
____def access_check(*args):
________sf = traceback.extract_stack()
________if len(sf) < 2 or sf[-2][2] != 'getMyClass':
____________raise "Access Denied"
________else:
____________return func(*args)
____return access_check

class MyClass:
____@classmethod
____def getMyClass(cls):
________return cls()

____@private
____def __init__(self):
________self.val = 9

if __name__ == '__main__':
____#a = MyClass()
____a = MyClass.getMyClass()
____print a.val

[xyzxyz1111]

为了更严密一点, 应该
<<<
if len(sf) < 2 or sf[-2][2] != 'getMyClass':
>>>
if len(sf) < 2 or sf[-2][2] != 'getMyClass' or sf[-2][0] != sf[-1][0]:

[CathySun118]

不了解ｐｙｔｈｏｎ