Python自学之乐-多继承顺序问题

最近自学Python到面向对象这里，感觉有必要把多继承这里好好理解一下，下面是我自己的一点小总结，如果有朋友觉得有什么不足的，还望不吝赐教！

1、#Author:clark

class Animal(object):#动物类    #kind = ""    def __init__(self,name,age,food):        self.Name = name        self.Age = age        self.food = food    def eat(self):        print("%s is eat %s" %(self.Name,self.food))    def barking(self):        print("%s is barking"%self.Name)class Relation(object):#动物关系类    def makefrined(self,obj):        print("%s is making friend with %s"%(self.Name,obj.Name))

class dog(Relation,Animal):#狗继承关系类与动物类    def __init__(self,name,age,food,kind):        super(dog,self).__init__(name,age,food)        self.kind = kind    def barking(self):        print("the barking of %s is wangwangwang!"%self.Name)class cat(Animal):    def __init__(self,name,age,food,kind):        super(cat,self).__init__(name,age,food)        self.kind = kind    def barking(self):        print("the barking of %s is miaomiaomiao"%self.Name)

dog_teddy = dog("ahuang",3,"bone","teddy")cat_xiaohua = cat("xiaohua",2,"fish","cat")dog_teddy.makefrined(cat_xiaohua)

运行结果为：ahuang is making friend with xiaohua一开始的疑惑在于，为什么类Relation的makefriend方法的，obj.name从哪来，怎么就运行成功了呢？后来发现Python的类继承的时候，有一个顺序：优先执行的是自己的构造方法，这里dog类有自己的构造方法，并且定义并继承了Animal类的name等属性，所以才会执行成功；后来我又想了想，如果把dog类自己的构造方法去掉呢？show me the code

2、#Author:clarkclass Animal(object):#动物类    #kind = ""    def __init__(self,name,age,food):        self.Name = name        self.Age = age        self.food = food    def eat(self):        print("%s is eat %s" %(self.Name,self.food))    def barking(self):        print("%s is barking"%self.Name)class Relation(object):#动物关系类    def makefrined(self,obj):        print("%s is making friend with %s"%(self.Name,obj.Name))

class dog(Relation,Animal):#狗继承关系类与动物类    # def __init__(self,name,age,food):    #     super(dog,self).__init__(name,age,food)    #     self.kind = kind    def barking(self):        print("the barking of %s is wangwangwang!"%self.Name)class cat(Animal):    def __init__(self,name,age,food,kind):        super(cat,self).__init__(name,age,food)        self.kind = kind    def barking(self):        print("the barking of %s is miaomiaomiao"%self.Name)

dog_teddy = dog("ahuang",3,"bone")cat_xiaohua = cat("xiaohua",2,"fish","cat")dog_teddy.makefrined(cat_xiaohua)

运行结果为：ahuang is making friend with xiaohua啊？为什么呢？明明dog没有构造方法了，按理说dog的实例对象去执行makefriend方法，发现类Relation没有name的属性，应该会抛错才对啊？才发现Python的类在继承的时候是去找构造方法的，这里dog没有构造方法，Relation也没有，那就找Animal的，发现有，就理所当然的执行成功了!其实大家可能知道python3类的继承顺序有一个从左往右的原则的，这里如果dog类的继承顺序写成dog(Animal,Relation)，可能会觉得好理解一点，其实从左往右是正确的，原则是找到构造方法就不再往下找，可以验证一下，show  me the code

3、#Author:clarkclass Animal(object):#动物类    #kind = ""    def __init__(self,name,age,food):        self.Name = name        self.Age = age        self.food = food    def eat(self):        print("%s is eat %s" %(self.Name,self.food))    def barking(self):        print("%s is barking"%self.Name)class Relation(object):#动物关系类    def __init__(self):#给Relation写一个构造方法        pass    def makefrined(self,obj):        print("%s is making friend with %s"%(self.Name,obj.Name))

class dog(Relation,Animal):#这里的顺序注意了，两个的类的不同顺序是有区别的    # def __init__(self,name,age,food):    #     super(dog,self).__init__(name,age,food)    #     self.kind = kind    def barking(self):        print("the barking of %s is wangwangwang!"%self.Name)class cat(Animal):    def __init__(self,name,age,food,kind):        super(cat,self).__init__(name,age,food)        self.kind = kind    def barking(self):        print("the barking of %s is miaomiaomiao"%self.Name)

dog_teddy = dog("ahuang",3,"bone")cat_xiaohua = cat("xiaohua",2,"fish","cat")dog_teddy.makefrined(cat_xiaohua)运行结果为报错：TypeError: __init__() takes 1 positional argument but 4 were given因为首先找到Relation的构造方法，发现没有属性，就不再往下找了，直接报错！可以试一下Animal写在前，就不会报错。

综合以上1、2、3我用自己的语言组织了下总结（供参考吐槽）：简单点就是Python的多继承有一个顺序，自己的构造方法 > 第一个类的构造方法 >大于第二个类的构造方法，以此类推，找到第一个构造方法为止，不再往下去找！！