python:__getattr__的合作超级调用

Question

我正在使用与此代码类似的somethign:

class BaseClass(object):
    def __getattr__(self, attr):
        return lambda:'1'

class SubClass(BaseClass):
    def foo(self):
        suffix = '2'
        return super(SubClass, self).foo() + suffix

class SubClass2(SubClass):
    def foo(self):
        suffix = '3'
        return super(SubClass2, self).foo() + suffix

o = SubClass2()
print o.foo()

我希望看到'123'的输出,但我得到一个错误AttributeError: 'super' object has no attribute 'foo'.Python甚至没有尝试使用基类__getattr__.

如果不修改基类,并保持两个超级调用相似,我就无法获得我想要的输出.有没有合适的超级电话模式在这里适合我？

我知道super()会以某种方式覆盖getattr以执行它需要做的事情,但是我想问是否有任何合理的解决方法允许__getattr__在适当的时候调用子类.

Answer 1

啊,这是一个很好的问题!

简而言之,这里发生的事情是CPython内部在进行属性查找时偶尔会采用快捷方式,而这种令人惊讶的行为是后果之一(另一个是提高性能).

为了准确了解在这种情况下发生了什么,我们需要冒险定义super:http: //hg.python.org/cpython/file/c24941251473/Objects/typeobject.c#l6689

特别注意它没有定义 tp_getattr (aka __getattr__),但确实定义了tp_getattro(aka __getattribute__):

PyTypeObject PySuper_Type = {
    PyVarObject_HEAD_INIT(&PyType_Type, 0)
    "super",                                    /* tp_name */
    ...
    0,                                          /* tp_getattr */
    ...
    super_getattro,                             /* tp_getattro */
    ...
};

(回想一下,__getattribute__被称为每次请求的属性,而不是__getattr__,它如果属性没有对象(大致上不存在只叫:如果属性不在对象 __dict__)).

接下来,查看super_getattro (aka super.__getattribute__)的定义 ,我们可以看到实现是 近似的:

class super(object):
    def __init__(self, obj_type, obj):
        self.obj_type = obj_type
        self.obj = obj

    def __getattribute__(self, attr):
        i = self.obj_type.__mro__.find(self.obj_type)
        i += 1
        while i < len(obj_type.__mro__):
            cur_type = self.obj_type.__mro__[i]
            cur_dict = cur_type.__dict___
            res = cur_dict.get(attr)
            if res is not None:
                return res
            i += 1
        return object.__getattribute__(self, attr)

这显然为什么super不能很好地使用__getattr__- super只检查父类中的属性' __dict__!

有趣的是:似乎pypy(从2.1.0开始)表现方式相同:

$ pypy super.py 
Traceback (most recent call last):
  File "app_main.py", line 72, in run_toplevel
  File "super.py", line 16, in <module>
    print o.foo()
  File "super.py", line 13, in foo
    return super(SubClass2, self).foo() + suffix
  File "super.py", line 8, in foo
    return super(SubClass, self).foo() + suffix
AttributeError: 'super' object has no attribute 'foo'

Answer 2

这似乎工作正常。我目前不明白为什么标准超类不这样做。

class super2(super):
    def __getattr__(self, attr):
        return self.__self__.__getattr__(attr)

class BaseClass(object):
    def __getattr__(self, attr):
        return lambda:'1'

class SubClass(BaseClass):
    def foo(self):
        suffix = '2'
        return super2(SubClass, self).foo() + suffix

class SubClass2(SubClass):
    def foo(self):
        suffix = '3'
        return super2(SubClass2, self).foo() + suffix

o = SubClass2()
print o.foo()