super() and @staticmethod interaction

The short answer to

Am I calling super(type) incorrectly here or is there something I'm missing?

is: yes, you're calling it incorrectly... AND (indeed, because) there is something you're missing.

But don't feel bad; this is an extremely difficult subject.

The documentation notes that

If the second argument is omitted, the super object returned is unbound.

The use case for unbound super objects is extremely narrow and rare. See these articles by Michele Simionato for his discussion on super():

• Things to Know About Python Super [1 of 3]
• Things to Know About Python Super [2 of 3] (this one specifically covers unbound super)
• Things to Know About Python Super [3 of 3]

Also, he argues strongly for removing unbound super from Python 3 here.

I said you were calling it "incorrectly" (though correctness is largely meaningless without context, and a toy example doesn't give much context). Because unbound super is so rare, and possibly just flat-out unjustified, as argued by Simionato, the "correct" way to use super() is to provide the second argument.

In your case, the simplest way to make your example work is

class First(object):  @staticmethod  def getlist():    return ['first']class Second(First):  @staticmethod  def getlist():    l = super(Second, Second).getlist()  # note the 2nd argument    l.append('second')    return la = Second.getlist()print a

If you think it looks funny that way, you're not wrong. But I think what most people are expecting when they see super(X) (or hoping for when they try it in their own code) is what Python gives you if you do super(X, X).

Since Second inherits form First, you can just use First.getlist() instead of passing in two arguments in super(i.e. super(Second, Second))

class First(object):   @staticmethod   def getlist():     return ['first']class Second(First):  @staticmethod  def getlist():    # l = super(Second, Second).getlist()    l = First.getlist()    l.append('second')    return la = Second.getlist()print (a)

When you call a normal method on a object instance, the method receives the object instance as first parameter. It can get the class of tte object and its parent class, so it makes sense to call super.

When you call a classmethod method on an object instance or on a class, the method receives the class as first parameter. It can get the parent class, so it makes sense to call super.

But when you call a staticmethod method, the method does not receive anything and has no way to know from what object or class it was called. That's the reason why you cannot access super in a staticmethod.