John Roth wrote:
"Tobiah" <[EMAIL PROTECTED]> wrote in message news:[EMAIL PROTECTED]

What is the purpose of the second argument to super()?


I've always  found the docs to be fairly confusing.
They didn't give me enough context to tell what
was going on. I also find the terminology confusing:
"type" seems to mean "new style class object", and
"object" seems to mean "instance."

I agree that the docs could probably do with some improvement here, but this is mostly because (I suspect) the type-based material has been shoehorned in to the existing documentation structure. My own suspicion was that a more radical revision would yield a better manual, but it doesn't look as though anybody has had time to attempt it.

Certainly super() is some of the deepest magic related to the new-style object hierarchy.

What happens with the second operand is a bit of
sleight of hand. The object returned from super()
gives you access to the methods on the next level up the
mro, however when you use it to invoke a method,
then the 'self' passed to that method is the second
object, not the instance returned from super().

This is a key point. I wonder whether we might use this thread to draft a patch to the docs for submission on SourceForge?

In most cases, this is exactly what you want, since if
the superclass method makes any changes to the
instance, you  want to be able to see them after the
call completes.

Quite. It's important that super() doesn't create a completely new object, because it's really about identifying an appropriate point in the inheritance hierarchy (method resolution order) and offering namespace access from that point up, rather than from the base instance given as the second argument to super(). This means that the same code can be included in many different namespace hierarchies and still work correctly in them all.

What is meant by the returning of an 'unbound' object
when the argument is omitted.


This is basically for calling static methods. Since a static
method is not passed an instance, you need a version of
the object returned from super() that doesn't bind the
method to an instance.

There is also the possibility that you might really want
to call an instance or class method as an unbound method,
explicitly passing it the instance. This is the reason that
the object returned from super() can't make the  distinction
automatically by simply checking for a static method.

Also, when would I pass an object as the second argument,
and when would I pass a type?


You need to pass the class object when you're calling
a class method. While __new__ is technically a static
method, for most practical purposes you can regard
it as a class method.

This is all good stuff. We should try to make sure the documentation gets enhanced.

regards
 Steve

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