It is perfectly explained in the standards here [1] saying that:

<quote>
In order to avoid infinite recursion in this method, its implementation should always call the base class method with the same name to access any attributes it needs, for example, object.__getattribute__(self, name).
</quote>

Therefore, I wrote a code following what the standard says:

<code>
class Sample():
def __init__(self):
self.a = -10

def __getattribute__(self, name):
if name == 'a':
return object.__getattribute__(self, name)

raise AttributeError()

s = Sample()
result = s.a
print(result)
</code>
I did not fall into recursion, and the output was
-10

I used here object.__getattribute__(self, name) cause the base class of Sample is object.

If I derive the Sample class from another class such as A, I should change object.__getattribute__(self, name) to A.__getattribute__(self, name) as the base class of class Sample is class A.

<code>
class A:
pass

class Sample(A):
def __init__(self):
self.a = -10

def __getattribute__(self, name):
if name == 'a':
return A.__getattribute__(self, name)

raise AttributeError()

s = Sample()

result = s.a
print(result)
</code>
which gives the same output as expected. No recursion and -10.

However, when I try the code without deriving from a class:

class AnyClassNoRelation:
pass

class Sample():
def __init__(self):
self.a = -10

def __getattribute__(self, name):
if name == 'a':
return AnyClassNoRelation.__getattribute__(self, name)

raise AttributeError()

s = Sample()

result = s.a
print(result)
and calling __getattribute__ via any class (in this example class AnyClassNoRelation) instead of object.__getattribute__(self, name) as the standard says call using the base class, I get the same output: no recursion and -10.

So my question:

How come this is possible (having the same output without using the base class's __getattribute__? Although the standards clearly states that __getattribute__ should be called from the base class.

<quote>
In order to avoid infinite recursion in this method, its implementation should always call the base class method with the same name to access any attributes it needs, for example, object.__getattribute__(self, name).
</quote>

Literally, I can call __getattribute__ with anyclass (except Sample cause it will be infinite recursion) I define and it works just fine. Could you explain me why that happens?

[1] https://docs.python.org/3/reference/datamodel.html#object.__getattribute__

On 24/05/2023 15:37, A KR wrote:
> It is perfectly explained in the standards here [1] saying that:
>
> <quote>
> In order to avoid infinite recursion in this method, its implementation should always call the base class method with the same name to access any attributes it needs, for example, object.__getattribute__(self, name).
> </quote>
>
> Therefore, I wrote a code following what the standard says:
>
> <code>
> class Sample():
> def __init__(self):
> self.a = -10
>
> def __getattribute__(self, name):
> if name == 'a':
> return object.__getattribute__(self, name)
>
> raise AttributeError()
>
> s = Sample()
> result = s.a
> print(result)
> </code>
> I did not fall into recursion, and the output was
> -10

While this works it's not how I understand the recommended pattern. I'd
rather treat "special" attributes first and then use the
__getattribute__ method of the base class as a fallback:

>> class Demo:
def __getattribute__(self, name):
if name == "answer":
return 42
return super().__getattribute__(name)

That way your special arguments,

>>> d = Demo()
>>> d.answer
42

missing arguments

>>> d.whatever
Traceback (most recent call last):
File "<pyshell#13>", line 1, in <module>
d.whatever
File "<pyshell#10>", line 5, in __getattribute__
return super().__getattribute__(name)
AttributeError: 'Demo' object has no attribute 'whatever'

and "normal" arguments are treated as expected

>>> d.question = "What's up?"
>>> d.question
"What's up?"

Eventual "special" arguments in the superclass would also remain accessible.

> However, when I try the code without deriving from a class:
>
> class AnyClassNoRelation:
> pass
>
> class Sample():
> def __init__(self):
> self.a = -10
>
> def __getattribute__(self, name):
> if name == 'a':
> return AnyClassNoRelation.__getattribute__(self, name)
>
> raise AttributeError()
>
> s = Sample()
>
> result = s.a
> print(result)
> and calling __getattribute__ via any class (in this example class AnyClassNoRelation) instead of object.__getattribute__(self, name) as the standard says call using the base class, I get the same output: no recursion and -10.
>
> So my question:
>
> How come this is possible (having the same output without using the base class's __getattribute__? Although the standards clearly states that __getattribute__ should be called from the base class.

AnyClassNoRelation does not override __getattribute__, so

>>> AnyClassNoRelation.__getattribute__ is object.__getattribute__
True

There is no sanity check whether a method that you call explicitly is
actually in an object's inheritance tree,

>>> class NoRelation:
def __getattribute__(self, name):
return name.upper()

>>> class Demo:
def __getattribute__(self, name):
return "<{}>".format(NoRelation.__getattribute__(self, name))

>>> Demo().some_arg
'<SOME_ARG>'

but the only purpose I can imagine of actually calling "someone else's"
method is to confuse the reader...
> <quote>
> In order to avoid infinite recursion in this method, its implementation should always call the base class method with the same name to access any attributes it needs, for example, object.__getattribute__(self, name).
> </quote>
>
> Literally, I can call __getattribute__ with anyclass (except Sample cause it will be infinite recursion) I define and it works just fine. Could you explain me why that happens?