I was reading official python documentation.
In the mentioned link, the second line states that:
Using this decorator requires that the class’s metaclass is ABCMeta or is derived from it.
But, I was successfully able to define the below given class.
from abc import abstractmethod
class A(object):
def __init__(self):
self.a = 5
@abstractmethod
def f(self):
return self.a
a = A()
a.f()
So, the code above worked fine. And also, I was able to create a subclass
class B(A):
def __init__(self):
super(B, self).__init__()
b = B()
b.f()
Without overriding the abstract method defined above.
So, basically does this mean that if my base class's metaclass is not ABCMeta(or derived from it), the class does not behave like an abstract class even though I have an abstract method in it?
That means, the documentation needs more clarity?
Or, is this behaviour useful somehow and I'm missing the point.
So, basically does this mean that if my base class's metaclass is not ABCMeta(or derived from it), the class does not behave like an abstract class even though I have an abstract method in it?
Correct.
All abstractmethod does is mark the method with __isabstractmethod__ = True. ABCMeta does all the real work. Here is the code for abstractmethod:
def abstractmethod(funcobj):
"""A decorator indicating abstract methods.
Requires that the metaclass is ABCMeta or derived from it. A
class that has a metaclass derived from ABCMeta cannot be
instantiated unless all of its abstract methods are overridden.
The abstract methods can be called using any of the normal
'super' call mechanisms.
Usage:
class C(metaclass=ABCMeta):
@abstractmethod
def my_abstract_method(self, ...):
...
"""
funcobj.__isabstractmethod__ = True
return funcobj