Question

我想为python类方法编写一个装饰器，它可以确定是从公共上下文还是私有上下文调用该方法。例如，给出以下代码

def public_check_decorator(f):
    def wrapper(self):
        if self.f is `called publicly`:  # <-- how do I make this line work correctly?
            print 'called publicly'
        else:
            print 'called privately'
        return f(self)
    return wrapper

class C(object):
    @public_check_decorator
    def public_method(self):
        pass

    def calls_public_method(self):
        self.public_method()

运行时执行最好看起来像这样：

>>> c = C()
>>> c.public_method()
called publicly

>>> c.calls_public_method()
called privately

有没有办法在python中这样做？也就是说，改变行

if self.f is `called publicly`:  # <-- how do I make this line work correctly?

给出所需的输出？

Answer 1

其中一些似乎试图对抗“蟒蛇”的潮流。这是合适的吗？

你知道双重非核心标准吗？它使方法“更私密”：

>>> class C(object):
...    def __hide_me(self):
...        return 11
...    def public(self):
...        return self.__hide_me()
...
>>> c = C()
>>> c.__hide_me()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
AttributeError: 'C' object has no attribute '__hide_me'
>>> c.public()
11
>>> c._C__hide_me()
11
>>>

那个私密吗？使用这种技术是pythonic。

Answer 2

鉴于包的名称决定是从私有上下文还是从公共上下文调用函数：

import inspect
import re


def run():
    package_name = '/my_package/'
    p = re.match(r'^.*' + package_name, inspect.stack()[0].filename).group()
    is_private_call = any(re.match(p, frame.filename) is not None for frame in inspect.stack()[1:])
    print(is_private_call)

尝试从包内运行，然后从包外面运行!!! 见inspect.stack()

在python中确定方法调用的范围/上下文

2 个答案: