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From: Ian Kelly <ian.g.kelly@gmail.com>
Newsgroups: comp.lang.python
Subject: Re: Where is decorator in this example code?
Date: Sat, 14 Nov 2015 09:08:33 -0700
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On Sat, Nov 14, 2015 at 7:46 AM, fl <rxjwg98@gmail.com> wrote:
> A following problem now is about the args in class decorate. I do not see
> args and kwargs are transferred by get_fullname(self).
>
>
> If I change
>
> return "<p>{0}</p>".format(func(*args, **kwargs))
>
> to
>
> return "<p>{0}</p>".format(func(*args))
>
> The outputs are the same.
>
> But it is quite different if it is changed to:
>
> return "<p>{0}</p>".format(func)

In this case you're not even calling the function, so the thing that
you're formatting in the string is the function object itself.

> What roles are args and kwargs? I know C language.
> For Python here, I don't see some rules on the args.

They are Python's version of "varargs" or variadic arguments.

If a function parameter declaration is prefixed with *, then that
parameter will collect all the remaining positional arguments. For
example, with the function declaration "def f(a, b, *c):", f may be
called with two or more arguments. The first two arguments will be
assigned to a and b, and c will be a tuple containing all the
remaining arguments. It is customary but not necessary to name this
parameter "args".

You can also do the inverse when calling a function. If x is a list or
tuple, then calling f(x) will pass the sequence x to f as a single
argument, while calling f(*x) will pass the contents of x as
individual arguments.

Similarly, if a function parameter declaration is prefixed with **,
then that parameter will collect all the keyword arguments that have
not been assigned to other parameters, in a dict. It is customary but
not necessary to name this parameter "kwargs". And again you can also
do the inverse when calling a function: if x is a dict, then calling
f(x) will pass along the dict as a single argument, while calling
f(**x) will pass the contents of the dict as individual keyword
arguments.

So if you have a function like this:

    def f(*args, **kwargs):
        return g(*args, **kwargs)

This collects all the arguments that were passed to f, and passes them
along to g in the same manner they were supplied to f.

In your example, removing **kwargs appeared to do nothing because no
keyword arguments were actually passed in.