calling super()

Wed Apr 4 16:08:57 EDT 2007

Jarek Zgoda wrote:
>> Hello, I have been trying to call the super constructor from my
>> derived class but its not working as expected. See the code:
>>
>> class HTMLMain:
>>     def __init__(self):
>>         self.text = "<HTML><BODY>";
>>         print(self.text);
>>     def __del__(self):
>>         self.text = "</BODY></HTML>";
>>         print(self.text);
>>
>> class NewPage(HTMLMain):
>>     def __init__(self):
>>         print 'derive2 init'
>>         super(NewPage, self).__init__();
>>     
>
> This should read: super(HTMLMain, self).__init__()
>   
Definitely, this is not true. Well, it depends what the OP wanted to do 
here, but in 99.9% of the cases, you want to use

class B(A):
    def method(self,*args):
       super(B,self).method(*args)

Look at here: http://www.python.org/doc/2.3.5/lib/built-in-funcs.html#l2h-66
It tells that super(type,[self]) will """Return the superclass of 
type.""". So super(B) will return the superclass of B, that is A. The 
built-in function "super" is very useful when you have diamond-shaped 
inheritance and you only want each inherited method to be called only 
once, IN THE CORRECT ORDER. If you only have single inheritance class 
trees, then super(B,self).method(*args) is identical to 
A.method(self,*args). You only need to worry about method calling order 
when you use multiple inheritance. However, using super is much nicer 
than calling the method of the base class directly, and it is 
syntactically cleaner, since you will only have a single reference to 
the base class, in the class definition header. (E.g. you can change the 
base class by replacing one word in the source code...)

Best,

  Laszlo