Question

有没有办法确定类属性是Property（__get__还是__set__？方法在 Determine if given class attribute is a property or not, Python object看起来只适用于property装饰器，这在我的情况下无效。

class Property(object):
    _value = None
    def __get__(self, instance, owner):
        return self._value 
    def __set__(self, instance, value):
        self._value = value * 2


class A(object):
    b = Property()

>>> a = A()
>>> type(A.p)
<type 'NoneType'>
>>> type(a.p)
<type 'NoneType'>

Answer 1

您的描述符返回None，因为它为类调用（instance属性在调用None时设置为__get__那个场景）。

你需要在没有调用描述符协议的情况下检索它，到达类__dict__属性是最直接的路径：

A.__dict__['p']

有关如何以及何时调用描述符的详细信息，请参阅Python Descriptor HOWTO。

或者，在property对象执行时执行，并在self设置为instance时返回None（因此在类上访问时）：

class Property(object): _value = None def __get__(self, instance, owner): if instance is None: return self return self._value def __set__(self, instance, value): self._value = value * 2

另见How does the @property decorator work?

演示：

>>> class Property(object): ... def __get__(self, instance, owner): ... return self._value ... def __set__(self, instance, value): ... self._value = value * 2 ... >>> class A(object): ... b = Property() ... >>> A.__dict__['b'] <__main__.Property object at 0x103097910> >>> type(A.__dict__['b']) <class '__main__.Property'> >>> class Property(object): ... _value = None ... def __get__(self, instance, owner): ... if instance is None: ... return self ... return self._value ... def __set__(self, instance, value): ... self._value = value * 2 ... >>> class A(object): ... b = Property() ... >>> A.b <__main__.Property object at 0x10413d810>

确定一个类属性是属性？

1 个答案: