1如果如图所示使用多重继承,我们将看到什么
2我们看到了基类被执行了两次Baseclass
3代码验证吧
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改进措施
5完美解决=基类执行了一次
6代码验证
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)
参考:本文参考学习《Python3 Object Oriented Programming》,根据自己理解改编,Dusty Phillips 著