1如果如圖所示使用多重繼承，我們將看到什麼

2我們看到了基類被執行了兩次Baseclass

3代碼驗證吧

class BaseClass:
    num_base_calls = 0
    def call_me(self):
        print("Calling method on Based Class")
        self.num_base_calls += 1
    
class LeftSubclass(BaseClass):
    num_left_calls = 0
    def call_me(self):
        BaseClass.call_me(self)
        print("Calling method on Left Subclass")
        self.num_left_calls += 1
        
class RightSubclass(BaseClass):
    num_right_calls = 0
    def call_me(self):
        BaseClass.call_me(self)
        print("Calling method on Right Subclass")
        self.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("Calling method on Subclass")
        self.num_sub_calls += 1

s = Subclass()
s.call_me() 
print(s.num_sub_calls, s.num_left_calls, s.num_right_calls, s.num_base_calls)

4改進措施

5完美解決=基類執行了一次

6代碼驗證

class BaseClass:
    num_base_calls = 0
    def call_me(self):
        print("Calling method on Based Class")
        self.num_base_calls += 1
    
class LeftSubclass(BaseClass):
    num_left_calls = 0
    def call_me(self):
        # BaseClass.call_me(self)
        super().call_me()
        print("Calling method on Left Subclass")
        self.num_left_calls += 1
        
class RightSubclass(BaseClass):
    num_right_calls = 0
    def call_me(self):
        # BaseClass.call_me(self)
        super().call_me()
        print("Calling method on Right Subclass")
        self.num_right_calls += 1

class Subclass(LeftSubclass, RightSubclass):
    num_sub_calls = 0
    def call_me(self):
        # LeftSubclass.call_me(self)
        # RightSubclass.call_me(self)
        super().call_me()
        print("Calling method on Subclass")
        self.num_sub_calls += 1

s = Subclass()
s.call_me() 
print(s.num_sub_calls, s.num_left_calls, s.num_right_calls, s.num_base_calls)

參考：本文參考學習《Python3 Object Oriented Programming》，根據自己理解改編，Dusty Phillips 著