多重继承－super使继承变得更清楚

多重继承的注意点，以及“下一个”的引用
详解super() 函数在多重继承中的作用

# 多重继承

# 多重继承的重复调用问题

class BaseClass:
    num_base_calls = 0
    def call_me(self):
        print("Calling method on BaseClass")
        self.num_base_calls += 1

class LeftSubClass(BaseClass):
    num_left_calls = 0
    def call_me(self):
        BaseClass.call_me(self)
        print("Calling methon on LeftC")
        self.num_left_calls += 1


class RightSubClass(BaseClass):
    num_right_calls = 0
    def call_me(self):
        BaseClass.call_me(self)
        print("Calling methon on RightC")
        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 SubC")
        self.num_sub_calls += 1

s = SubClass()
print(s.call_me())
# Calling method on BaseClass
# Calling methon on LeftC
# Calling method on BaseClass
# Calling methon on RightC
# Calling method on SubC
print(s.num_base_calls, s.num_left_calls, s.num_right_calls, s.num_sub_calls)
# 2 1 1 1

# 由上面可知，我们的base类的call_me方法被调用了两次，是绝对不允许的！


# 利用super函数实现调用“下一个方法”

class BaseClass:
    num_base_calls = 0
    def call_me(self):
        print("Calling method on BaseC")
        self.num_base_calls += 1

class LeftSubClass(BaseClass):
    num_left_calls = 0
    def call_me(self):
        super().call_me()
        print("Calling method on LeftC")
        self.num_left_calls += 1

class RightSubClass(BaseClass):
    num_right_calls = 0
    def call_me(self):
        super().call_me()
        print("Calling method on RightC")
        self.num_right_calls += 1

class SubClass(LeftSubClass, RightSubClass):
    num_sub_calls = 0
    def call_me(self):
        super().call_me()
        print("Calling method on SubC")
        self.num_sub_calls += 1

e = SubClass()
e.call_me()
# Calling method on BaseC
# Calling method on RightC
# Calling method on LeftC
# Calling method on SubC
print(e.num_base_calls, e.num_left_calls, e.num_right_calls, e.num_sub_calls)
# 1 1 1 1

# 注释：与直接调用超类中的方法不同的是，super()函数调用的对于
# 具有多个超类中，如果有多个相同的方法（比如这里，基类和同等级的类中都有
# 相同的方法，那么，super函数调用的是同等级类中的方法，也就是俗称的“下一个”方法

# 从上面和这里可以看出，super函数实例化的一个对象是“下一个”类的实例化