如何将实例属性标记为未在基类中实现? (与 this 问题不同,该问题讨论将 class 属性标记为未实现,但也许我没有正确理解基类...)
例如我想要类似的东西
class Base():
def __init__(self):
self.x = NotImplemented
class GoodSub(Base):
def __init__(self, x):
super().__init__()
self.x = x #good
class BadSub(Base):
def __init__(self):
super().__init__()
#forgot to set self.x
good = GoodSub(5)
bad = BadSub(-1)
good.x #returns 5
bad.x #throws error because x not implemented
或者,是否有更好的方法来强制 Base
的所有子类在初始化时设置 self.x
属性?
最佳答案
一种带有类装饰器和描述符的解决方案(__get__
方法):
def abstract_variables(*args):
class av:
def __init__(self, error_message):
self.error_message = error_message
def __get__(self, *args, **kwargs):
raise NotImplementedError(self.error_message)
def f(klass):
for arg in args:
setattr(klass, arg, av('Descendants must set variable `{}`'.format(arg)))
return klass
return f
@abstract_variables('x', 'y')
class Base:
def __init__(self):
pass
class Derived(Base):
x = 10
b = Base()
d = Derived()
print(d.x) # prints 10
print(d.y) # raises NotImplementedError
打印:
10
Traceback (most recent call last):
File "main.py", line 28, in <module>
print(d.y)
File "main.py", line 7, in __get__
raise NotImplementedError(self.error_message)
NotImplementedError: Descendants must set variable `y`
关于实例属性的 Python NotImplementedError,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/56997102/