python - 是否允许在抽象类中使用用户定义的装饰器？或者说，应该继承后使用吗？

Question

例如，假设我定义了一个装饰器，名为:decorate

def decorate(func):
  def inside_func(*args, **kwargs):
    # Do something
    return func(*args, **kwargs)
  return inside_func

接下来，假设我正在编写一个名为 Model 的抽象类

from abc import ABC, abstractmethod

class Model(ABC):
  def __init__(self, value):
      self.value = value
      super().__init__()

  @abstractmethod
  @decorate # <-------------------- IS @decorate ALLOWED HERE?
  def do_something(self):
      pass

或者，应该是:

from abc import ABC, abstractmethod

class Model(ABC):
  def __init__(self, value):
    self.value = value
    super().__init__()

  @abstractmethod
  def do_something(self):
    pass

# Inherit the base class
class DoAdd42(Model):
   @decorate # <----------------------- SHOULD @decorate BE HERE INSTEAD?
   def do_something(self):
     return self.value + 42

如果两者都允许，是否有“最佳实践”方法来做到这一点？

Answer 1

两者都允许。因为abstractmethod，你的类中的decorate和something都是函数。您可以将它们打印到控制台进行验证。

In [2]: def decorate(func):
   ...:   def inside_func(*args, **kwargs):
   ...:     # Do something
   ...:     return func(*args, **kwargs)
   ...:   return inside_func
   ...:
In [7]: from abc import ABC, abstractmethod
   ...:
   ...: class Model(ABC):
   ...:   def __init__(self, value):
   ...:       self.value = value
   ...:       super().__init__()
   ...:
   ...:   @abstractmethod
   ...:   @decorate # <-------------------- IS @decorate ALLOWED HERE?
   ...:   def do_something(self):
   ...:       pass
   ...:   # even no self is valid for class in Python3+
   ...:   def whatever():
   ...:       print('fff')
   ...:

In [8]: print(abstractmethod)
<function abstractmethod at 0x100ff86a8>

In [9]: print(Model.do_something)
<function decorate.<locals>.inside_func at 0x1041031e0>

In [10]: print(Model.whatever)
<function Model.whatever at 0x103b48950>

In [11]: Model.whatever()
fff

我不熟悉abstractmethod，因此我不知道最佳实践以及有关其规则的更多详细信息。你可以根据我给你的信息，结合你自己对abstractmethod的理解，尝试做出自己的判断。