[Python-3000] Fwd: defop ?

[Guido van Rossum]

On 11/25/06, Calvin Spealman <ironfroggy at gmail.com> wrote:
> I will make only one more comment and then ill drop my comments
> without direct questions.
>
> On 11/25/06, Guido van Rossum <guido at python.org> wrote:
> > Hm. The double colon rubs me the wrong way (Perl and/or C++). But
> > apart from that, if this is the solution, I'm not sure the problem
> > you're trying to solve is really worth solving. I just don't expect
> > there will be all that many generic operations that need to be stuffed
> > into arbitrary classes. Maybe I'll just take back what I said about
> > wanting all such operations to be inside the class. Or maybe I'd be
> > happier if there was a decorator indicating the name of the operation.
>
> I don't care about the syntax. anything that can denote an expression
> (the left of the ::) and a name (the right of the ::) is OK and I dont
> care how its denoted.
>
> > Also, I think you're overloading 'def' in a non-obvious way.
> > Currently, "def foo..." means an assignment to the local variable foo.
> > I would expect that if we extend the syntax for the thing between
> > 'def' and the argument list to be more than just a name, it should
> > still be considered an assignment target. But that's clearly not what
> > you're after. From that POV, I find defop (while still unpleasant for
> > other reasons) more "honest" than your overloading of def -- at least
> > defop says upfront that it's not just an assignment. (Although the
> > similarity with def is still confusing IMO.)
>
> You must be misunderstanding me. I am not saying that its not an
> assignment. It would not change what def really means. operator::len
> would be the actual name of the function to be created and the name of
> the global, local, or class attribute it is bound to. I am saying
> operator::len would become something like
> MyClass.__dict__[operator::len] and what operator::len evaluates to, i
> dont know. something that represents what it is. Maybe just a tuple. I
> would expect it also exist for any assignment. special casing being
> bad and all.

Do you realize that expr[expr::expr] already has a meaning?

You seem to be doing an exceptionally poor job explaining your
proposal; you are introducing a new token "::" without ever explaining
what you propose it should mean. How does operator::len differ from
operator.len?

> > Still not convinced? Focus on other problems first. This isn't the
> > most important problem we're trying to solve.
> >
> > PS, It's __builtin__, not __builtins__ -- the latter is an
> > unfortunately named but ultimately unimportant implementation detail
> > (unimportant unless you're implemented restricted python, that is);
> > the former is the module that is implied at the end of every free name
> > search.
> >
> >
> > --Guido
>
> --
> Read my blog! I depend on your acceptance of my opinion! I am interesting!
> http://ironfroggy-code.blogspot.com/
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-- 
--Guido van Rossum (home page: http://www.python.org/~guido/)