How does calling super() with the 2-argument form differ from directly referencing the method and passing in "self" manually?

Question

A bit of an odd question, consider the following code:

class A: def mymethod(self): return "mymethod from class A" class B(A): def mymethod(self): return "mymethod from class B" class C(B): # in this class I want "mymethod" to return the same string returned from mymethod in the A class, with 2 exclamation points (!!) appended # obviously this is a simplified example with dumb logic def mymethod(self): ss = super(B, self).mymethod() # OR ss = A.mymethod(self) return ss + " !!"

I am fairly inexperienced when it comes to OOP so excuse me if the answer is extremely obvious.

As far as I can see, there is no difference between these 2 approaches, and as such I fail to understand how super() can be useful inside class methods.

Barmar · Accepted Answer · 2020-10-14 20:31:53Z

2

Using super(B, self) removes a level of dependency. If the definition of B is changed to use a different superclass, you'll automatically call the method from that class. If you hard-code A.mymethod(self), you'll need to update that when B is redefined.

answered Oct 14, 2020 at 20:31

Barmar

789k57 gold badges554 silver badges669 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

bool3max Over a year ago

OK, that clears up why super() comes in handy, thanks. But I am right in that there is no functional difference between the 2 methods so long as the code does not change?

Barmar Over a year ago

Yes, they're the same in this instance.

Collectives™ on Stack Overflow

How does calling super() with the 2-argument form differ from directly referencing the method and passing in "self" manually?

1 Answer 1

2 Comments

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

2 Comments

Related