Mixin 类来跟踪属性请求 - __attribute__ 递归

Question

我正在尝试创建一个必须是其他类的超类的类，跟踪它们的属性请求。我想过使用“getattribute”来获取所有属性请求，但它会生成递归：

class Mixin(object):
 def __getattribute__ (self, attr):
  print self, "getting", attr
         return self.__dict__[attr]

我知道为什么我会得到递归：它是为了 self.dict 调用，它让人想起 > 递归获取属性。我尝试按照其他帖子中的建议更改 "return object.__getattribute__(self,attr)" 中的最后一行，但会调用递归。

Answer 1

试试这个：

class Mixin(object):
    def __getattribute__ (self, attr):
        print self, "getting", attr
        return object.__getattribute__(self, attr)

如果您仍然遇到递归问题，这是由您没有向我们展示的代码引起的

>>> class Mixin(object):
...     def __getattribute__ (self, attr):
...         print self, "getting", attr
...         return object.__getattribute__(self, attr)
...
>>> Mixin().__str__
<__main__.Mixin object at 0x00B47870> getting __str__
<method-wrapper '__str__' of Mixin object at 0x00B47870>
>>> Mixin().foobar
<__main__.Mixin object at 0x00B47670> getting foobar
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 4, in __getattribute__
AttributeError: 'Mixin' object has no attribute 'foobar'
>>>

，这是与 Bob 的 Mylist 结合使用的结果

>>> class Mylist(Mixin):
...     def __init__ (self, lista):
...         if not type (lista) == type (""):
...             self.value = lista[:]
...     def __add__ (self,some):
...         return self.value + some
...     def __getitem__ (self,item):
...         return self.value[item]
...     def __getslice__ (self, beg, end):
...         return self.value[beg:end]
...
>>> a=Mylist([1,2])
>>> a.value
<__main__.Mylist object at 0x00B47A90> getting value
[1, 2]

Answer 2

这是代码：

from Es123 import Mixin
class Mylist(Mixin):
    def __init__ (self, lista):
        if not type (lista) == type (""):
            self.value = lista[:]
    def __add__ (self,some):
        return self.value + some
    def __getitem__ (self,item):
        return self.value[item]
    def __getslice__ (self, beg, end):
        return self.value[beg:end]
a = Mylist ([1,2])
a.value

然后python返回“RuntimeError：超出最大递归深度”