python - 如何确定描述符的类别？

Question

在此示例代码中，我想确定 x 是否是 TestProperty 的实例:

class TestProperty(object):
    def __init__(self, name):
        self._name = name

    def __get__(self, instance, cls):
        return getattr(instance, self._name)

    def __set_(self, instance, value):
        setattr(instance, self._name, value)

class Test(object):
    x = TestProperty("x")

print isinstance(Test.x, TestProperty)

但是，我遇到以下异常:

Traceback (most recent call last):
  File "/home/zenoss/testproperties.py", line 14, in <module>
    print isinstance(Test.x, TestProperty)
  File "/home/zenoss/testproperties.py", line 6, in __get__
    return getattr(instance, self._name)
AttributeError: 'NoneType' object has no attribute 'x'

当属性是描述符时，是否有办法判断它是否是类的实例？

Answer 1

使用当前的__get__，Test.x会导致AttributeError，因为当代码使用类、instance<访问描述符时 被传递None； (=> getattr(None, 'x') => None.x)

您应该修改__get__来处理这种情况:

>>> class TestProperty(object):
...     def __init__(self, name):
...         self._name = name
...     def __get__(self, instance, cls):
...         if instance is None:  # To handle access through class, not instance
...             return self       # returns the TestProperty instance itself.
...         return getattr(instance, self._name)
...     def __set_(self, instance, value):
...         setattr(instance, self._name, value)
... 
>>> class Test(object):
...     x = TestProperty("x")
... 
>>> isinstance(Test.x, TestProperty)
True

<小时/>

顺便说一句，您可能知道，使用 x = TestProperty("x")，通过实例访问 x 属性将导致另一个异常，因为它将调用__get__ (-> getattr(..) -> __get__ -> getattr(..) -> ...) 递归直到堆栈溢出。