类方法，继承小问题

sunmichace

1. #类方法：
   
2. class Spam:
   
3.     numInstance = 0
   
4.     def __init__(self):
   
5.         Spam.numInstance += 1
   
6.     def printNumInstance(aclass):
   
7.         print('Number of instances:', aclass.numInstance, aclass)
   
8.     printNumInstance = classmethod(printNumInstance)
   
9. 
   
10. class Sub(Spam):
    
11.     def printNumInstance(aclass):
    
12.         print('Extra stuff ...', aclass)
    
13.         Spam.printNumInstance()
    
14.     printNumInstance = classmethod(printNumInstance)
    
15. 
    
16. class Other(Spam):pass
    
17. 
    
18. >>> x, y = Sub(), Spam()
    
19. >>> x.printNumInstance()
    
20. Extra stuff ... <class '__main__.Sub'>
    
21. Number of instances: 2 <class '__main__.Spam'>
    
22. >>> Sub.printNumInstance()
    
23. Extra stuff ... <class '__main__.Sub'>
    
24. Number of instances: 2 <class '__main__.Spam'>
    
25. >>> y.printNumInstance()
    
26. Number of instances: 2 <class '__main__.Spam'>
    
27. >>> z = Other()
    
28. >>> z.printNumInstance()
    
29. Number of instances: 3 <class '__main__.Other'>
    
30. #x,z 均继承自Spam，为什么调用Spam.printNumInstance()时，
    
31. #aclass返回<class '__main__.Spam'>,而另一个却返回<class '__main__.Other'>

复制代码

x,z 均继承自Spam，为什么调用Spam.printNumInstance()时，aclass返回<class '__main__.Spam'>,而另一个却返回<class '__main__.Other'>

醉酒青牛

x和z分别是类Sub和Other的实例化对象。而Sub和Other类均是类Spam的子类。
但是，Sub类中重新定义了printNumInstance函数，且其函数体内由对父类函数printNumInstance的调用语句
Spam.printNumInstance()，因此x.printNumInstance()会返回Number of instances: 2 <class '__main__.Spam'>。
而对于Other类，其类内并没有定义任何函数，所以会直接继承父类Spam的printNumInstance的函数，当其调用z.printNumInstance()时，会将z所述类Other输出，即Number of instances: 3 <class '__main__.Other'>。