为什么在描述符类的__get__中需要“if instance is None”？

Question

我从Effective Python第31项获得以下示例：

from weakref import WeakKeyDictionary
class Grade(object):
    def __init__(self):
        self._values = WeakKeyDictionary()
    def __get__(self, instance, instance_type):
        if instance is None: return self
        return self._values.get(instance, 0)

    def __set__(self, instance, value):
        if not (0 <= value <= 100):
            raise ValueError('Grade must be between 0 and 100')
        self._values[instance] = value


# Example 16
class Exam(object):
    math_grade = Grade()
    writing_grade = Grade()
    science_grade = Grade()

first_exam = Exam()
first_exam.writing_grade = 82
second_exam = Exam()
second_exam.writing_grade = 75
print('First ', first_exam.writing_grade, 'is right')
print('Second', second_exam.writing_grade, 'is right')

我无法想出在if instance is None: return self中有__get__的任何理由。 Exam（或使用Grade）实例的其他潜在类如何成为None？

Answer 1

当访问类的描述符时，Python将为实例传递None。

在这种情况下，通过返回self，您可以访问类上的描述符对象，而不必绕过协议（通过访问ClassObj.__dict__['name_of_descriptor']）。

>>> class DemoDescriptor:
...     def __get__(self, instance, type_):
...         if instance is None:
...             print('Accesing descriptor on the class')
...             return self
...         print('Accessing descrtor on the instance')
...         return 'Descriptor value for instance {!r}'.format(instance)
... 
>>> class DemoClass(object):
...     foo = DemoDescriptor()
... 
>>> DemoClass.foo  # on the class
Accesing descriptor on the class
<__main__.DemoDescriptor object at 0x1041d3c50>
>>> DemoClass.__dict__['foo']  # bypassing the descriptor protocol
<__main__.DemoDescriptor object at 0x1041d3c50>
>>> DemoClass().foo  # on the instance
Accessing descrtor on the instance
'Descriptor value for instance <__main__.DemoClass object at 0x1041d3438>'

对于您的具体情况，Exam.math_grade，Exam.writing_grade或Exam.science_grade中的每一个都会调用Grade.__get__，为实例传递None，{{1对于Exam。