CODEWITHSUNDEEP
pythonclass
run_the_race
run_the_race

Reputation: 2418

Python refer to class of current code (not extended or inherited class)

In the below code I have to refer to Foo a second time (i.e. in super(Foo, self) Is there anything smarter I can insert, such that if I rename Foo, the other smarter tag doesnt need to be updated? It does not seem like very DRY code.

Note: I have to specify the starting point for super, the arg can't just be left out because this class gets extended.

class Foo(Bar):
    def __init__(self, *args, **kwargs):
        super(Foo, self).__init__(*args, **kwargs)

Upvotes: 0

Views: 59

Answers (1)

felipe
felipe

Reputation: 8035

super() without arguments in Python 3 is the equivalent of super(Foo, self) in your case, and based on your bold description it is also expected to work- even if Foo is a parent in subsequent code.

class Foo(Bar):
    def __init__(self, *args, **kwargs):
        super().__init__(*args, **kwargs)

Should work fine. There are three ways of using super():

class Parent:
    def __init__(self, a):
        self.a = a
        
class Child(Parent):
    def __init__(self, a, b):
        super(Child, self).__init__(a)    
        self.b = b    

class ChildSuper(Parent):
    def __init__(self, a, b):
        super().__init__(a)
        self.b = b
        
class ChildClass(Parent):
    def __init__(self, a, b):
        Parent.__init__(self, a)
        self.b = b
        
assert Child(1, 2).a == 1
assert ChildSuper(1, 2).a == 1
assert ChildClass(1, 2).a == 1

Upvotes: 1

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