python_多重继承

最新推荐文章于 2023-09-24 12:18:50 发布

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最新推荐文章于 2023-09-24 12:18:50 发布

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分类专栏： python

本文链接：https://blog.youkuaiyun.com/Folivoraxue/article/details/104212817

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本文深入探讨了Python多重继承中出现的钻石问题，通过具体代码示例，展示了如何使用super()来正确调用基类的方法，避免重复调用，确保方法调用的顺序遵循C3线性化原则。

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钻石的问题

class BaseClass:
    num_base_calls = 0
    def call_me(self):
        print("Base Class")
        BaseClass.num_base_calls += 1
        
class LeftSubclass(BaseClass):
    num_left_calls = 0
    def call_me(self):
        BaseClass.call_me(self)
        print("Left Class")
        LeftSubclass.num_left_calls += 1
        
class RightSubclass(BaseClass):
    num_right_calls = 0
    def call_me(self):
        BaseClass.call_me(self)
        print("Right Class")
        RightSubclass.num_right_calls += 1
        
class Subclass(LeftSubclass,RightSubclass):
    num_sub_calls = 0
    def call_me(self):
        LeftSubclass.call_me(self)
        RightSubclass.call_me(self)
        print("Sub Class")
        Subclass.num_sub_calls += 1

执行结果：

C:\Users\39621>python -i d:\2019\py1.py
>>> s=Subclass()
>>> s.call_me()
Base Class
Left Class
Base Class
Right Class
Sub Class

BaseClass的call_me方法被调用了两次。
解决方法super().call_me()

class BaseClass:
    num_base_calls = 0
    def call_me(self):
        print("Base Class")
        BaseClass.num_base_calls += 1
        
class LeftSubclass(BaseClass):
    num_left_calls = 0
    def call_me(self):
        super().call_me()
        print("Left Class")
        LeftSubclass.num_left_calls += 1
        
class RightSubclass(BaseClass):
    num_right_calls = 0
    def call_me(self):
        super().call_me()
        print("Right Class")
        RightSubclass.num_right_calls += 1
        
class Subclass(LeftSubclass,RightSubclass):
    num_sub_calls = 0
    def call_me(self):
        super().call_me()
        print("Sub Class")
        Subclass.num_sub_calls += 1

执行结果：

C:\Users\39621>python -i d:\2019\py1.py
>>> s=Subclass()
>>> s.call_me()
Base Class
Right Class
Left Class
Sub Class

Subclass的call_me()调用了super().call_me()，其实就是引用了LeftSubclass.call_me();
然后LeftSubclass.call_me()调用了super().call_me()，但这时super()引用了RightSubclass.call_me()