Python\uuuGetAttribute\uuuuuu回退到\uuuuGetAttr__

我遇到一种情况，getattribute回退到getattr，然后再次调用getattribute

如何再次调用当前的getattribute？我很困惑

class Count(object):

    def __init__(self,mymin,mymax):
        self.mymin=mymin
        self.mymax=mymax
        self.current=None

    def __getattr__(self, item):
            print("akhjhd")
            self.__dict__[item]=0
            return 0

    def __getattribute__(self, item):
        print("this is called first")
        if item.startswith('cur'):
            print("this raised an error")
            raise AttributeError
        print("This will execute as well")
        return object.__getattribute__(self,item)


obj1 = Count(1,10)
print(obj1.mymin)
print(obj1.mymax)
print(obj1.current)

控制台输出：

this is called first
This will execute as well
1
this is called first
This will execute as well
10
this is called first
this raised an error
akhjhd
this is called first
This will execute as well
0

---

• 调用

  getattr

  是因为

  getattribute

  引发

  AttributeError

• self.\uuuu dict\uuuu

  调用对

  getattribute

清洁代码并添加

打印（项目）

以使其更清晰：

class Count(object):
    def __init__(self):
        self.current = None

    def __getattr__(self, item):
        print("in getattr")
        self.__dict__[item] = 0
        return 0

    def __getattribute__(self, item):
        print(item)
        print("in __getattribute__ 1")
        if item.startswith('cur'):
            print("starts with 'cur'")
            raise AttributeError
        print("in __getattribute__ 2")
        return object.__getattribute__(self, item)


obj1 = Count()
print(obj1.current)

输出

current
in __getattribute__ 1
starts with 'cur'
in getattr
__dict__
in __getattribute__ 1
in __getattribute__ 2
0

---

• 调用

  getattr

  是因为

  getattribute

  引发

  AttributeError

• self.\uuuu dict\uuuu

  调用对

  getattribute

清洁代码并添加

打印（项目）

以使其更清晰：

class Count(object):
    def __init__(self):
        self.current = None

    def __getattr__(self, item):
        print("in getattr")
        self.__dict__[item] = 0
        return 0

    def __getattribute__(self, item):
        print(item)
        print("in __getattribute__ 1")
        if item.startswith('cur'):
            print("starts with 'cur'")
            raise AttributeError
        print("in __getattribute__ 2")
        return object.__getattribute__(self, item)


obj1 = Count()
print(obj1.current)

输出

current
in __getattribute__ 1
starts with 'cur'
in getattr
__dict__
in __getattribute__ 1
in __getattribute__ 2
0

---

您需要咨询python

摘录如下：

无条件调用以实现类实例的属性访问。如果该类还定义了

\uuuuuGetAttr\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu（）

，则后者将不会被调用，除非

\uuuuuuuuuuuuuuuuuuuuuuuuGetAttribute\uuuuu

我在你的代码中看到：

if item.startswith('cur'):
    print("this raised an error")
    raise AttributeError

所以我认为您是故意这么做的
您需要咨询python

摘录如下：

无条件调用以实现类实例的属性访问。如果该类还定义了
\uuuuuGetAttr\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu（）
，则后者将不会被调用，除非
\uuuuuuuuuuuuuuuuuuuuuuuuGetAttribute\uuuuu

我在你的代码中看到：

if item.startswith('cur'):
    print("this raised an error")
    raise AttributeError

所以我认为你是故意这么做的
getattr
之所以被调用，是因为
getattribute
引发了
AttributeError
同意，但再次调用了，为什么？因为
self.\uuuu dict\uuuuu
第二次调用它，我添加了一个示例……感谢DeepSpace
getattr
之所以调用它，是因为
getattribute
引发了
AttributeError
同意，但再次调用了\uuuuuuuu\getattribute\uu，为什么会这样？因为我第二次添加答案时，self.\uuuu dict\uuuu.\uuu
调用了它，并添加了一个示例……感谢Deepspace如果我没有getattr方法并在getattribute中返回，为什么getattribute中的第一次打印会被一次又一次地打印（意味着getattribute的递归）……这怎么会不止一次被呼叫？@Pravin我无法重现这种行为。如果我注释掉了
getattr
，那么
getattribute
中的
AttributeError
就会传播。如果我没有getattr方法并在getattribute中返回，为什么在getattribute中第一次打印会被反复打印（getattribute的平均递归）……这怎么会不止一次被呼叫？@Pravin我无法重现这种行为。如果我将
getattr
注释掉，则
getattribute
中的
AttributeError
将传播。