Mixin类跟踪属性请求 - __attribute__递归

Question

我想创建一个类，必须是其他类的超类，跟踪他们的属性请求。我想用“的getAttribute”，这得到所有属性的请求，但它产生的递归：

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

我知道为什么我得到递归：它是自我。 字典调用哪个回调getattribute递归。我试图改变"return object.__getattribute__(self,attr)"的最后一行，就像其他帖子中的建议一样，但是递归被调用。

Answer 1

试试这个：

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

如果你仍然得到递归的问题，它是由代码导致你还没有告诉我们

>>> 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' 
>>> 

这里是结果，当Bob的Mylist合并

>>> 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 

那么Python返回 “RuntimeError：最大递归深度超过”