python - 如何使每个类的类属性分开以避免被不同类覆盖？

Question

简单示例 Parent 是具有 version 属性的类。我想从此类继承，但避免在所有继承类之间共享版本属性。

class Parent(object):
    versions = []

    @classmethod
    def add_version(cls, version):
        cls.versions.append(version)

class P1(Parent):
    pass

class P2(Parent):
    pass

class Thing(object):
    pass

P1.add_version(Thing)
# should be Thing
print(P1.versions)
# should be empty [] by my design but it works other way in Python
# how to achieve to different versions fields without coding it in child classes?
print(P2.versions)

由于类之间共享版本而产生这样的结果 - 如何避免？

[<class '__main__.Thing'>]
[<class '__main__.Thing'>]

Answer 1

从技术上讲，您不需要一个元类——一个简单的类装饰器也可以工作(好吧，有点……):

def versions_container(cls):
    if "versions" not in cls.__dict__:
        print("adding versions to class {}".format(cls.__name__))
        cls.versions = []
    return cls


@versions_container # this one is useless but anyway...
class Parent(object):
    versions = []
    @classmethod
    def add_version(cls, version):
        cls.versions.append(version)

@versions_container
class P1(Parent):
    pass

@versions_container
class P2(Parent):
    pass

class Thing(object):
    pass

P1.add_version(Thing)
print(P1.versions)
print(P2.versions)

但是这并不比在每个子类中手动添加versions 属性好多少。

自定义元类解决方案并不是什么大问题，它可以确保您的代码的用户不会违反预期:

class VersionsContainer(type):
    def __new__(meta, name, bases, attrs):
        if "versions" not in attrs:
            attrs["versions"] = []
        return type.__new__(meta, name, bases, attrs)

class Parent(object, metaclass=VersionContainer):
    @classmethod
    def add_version(cls, version):
        cls.versions.append(version)

class P1(Parent):
    pass

class P2(Parent):
    pass

class Thing(object):
    pass

P1.add_version(Thing)
print(P1.versions)
print(P2.versions)