在 python 中将实例类型更改为继承

Question

我有一个 class Person 和 2 classes Child 和 Adult 继承自 `Person.


class Person:

    def __init__(self, name):
        self.name = name
        self.age = 0

    def say_hi(self):
        print("Hi, I am " + self.name)

    def set_age(self, age):
        self.age = age

class Child(Person):

    def play_with_children(self):
        print("Playing with children")

class Adult(Person):

    def drink_alcohol(self):
        print("drinking")

我想从 Person 创建一个实例然后将它设置为 age 根据它，这个实例应该 "become" a Child 或 Adult.

此外，如果它是 Child 并且增加了 age，我希望它增加到 "become" 和 Adult。请注意，此代码有一个错误，因为 Person 的构造函数需要一个 name，因为它旨在创建一个新实例。

person1 = Person("Marvin")

person1.set_age(15)

if person1.age < 21:
    person1 = Child()# This is an error => expects the name
    #But I wouldn't like to handle all members like this => person1 = Child(person1.name)
    #I don't want a new instance, is that even posible?

person1.say_hi() 
person1.play_with_children()


person1.set_age(25)

if person1.age > 21:
    person1 = Adult()# This is an error => expects the name
    #But I wouldn't like to handle all members like this => person1 = Adult(person1.name)
    #I don't want a new instance, is that even posible?

person1.say_hi()   
person1.drink_alcohol()

这可能吗？
是否可以在不创建新实例的情况下完成？

PS：这只是一个示例代码，显示了我遇到的一个更复杂的问题（不是真正的儿童和成人：D）。

Answer 1

如评论中所述，在 Python 3 中我们可以访问任何实例的 __class__ 并更改它。

虽然你必须小心，但你可以将几个类的属性添加到同一个实例并创建一个奇怪的怪物（方法是安全的）。

person1 = Person("Marvin")

person1.set_age(15)

if person1.age < 21:
    person1.__class__ = Child

person1.say_hi() 
person1.play_with_children()


person1.set_age(25)

if person1.age > 21:
    person1.__class__ = Adult


person1.say_hi()   
person1.drink_alcohol()

所以怪物：

class Dog:

    def wouf(self):
        print("wouf")

person1.__class__ = Dog
print("Attributes are saved, name: "+ person1.name)
#But methods not, so now I can't use person1.say_hi()
person1.wouf()

在 python 中将实例类型更改为继承

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