Second argument to super().

[John Roth]

"Tobiah" <toby at rcsreg.com> wrote in message 
news:1110396167.53bd67fed97c9e826a92b284283b1455 at teranews...
> 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."

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().

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.

> 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.

John Roth
>
> Thanks,
>
> Tobiah