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comp.lang.ruby

Re: syntax sugar: treating an object like a method

Eric Mahurin

5/18/2005 10:05:00 PM

> >>> I'm thinking of yet another RCR and would like to see if
> anybody
> >>> likes it. I think this would be a sweet feature. Here are
> the
> >>> basic components:
> >> Please don't. This would be awful. Ruby doesn't need a ()
> >> operator.
> > Oh, never say never. I still fantasize about first-order
> > functions/ methods (although that need would probably be
> better
> > served by other means). For this particular purpose I would
> tend
> > to agree, though.
>
> As I understand it, Ruby already has first-order functions;
> they're
> just not (currently) callable with foo(). Ruby 1.9 has an
> experimental change that does this for local variables that
> have
> Methods or Procs in them.

So procs and methods are getting the () operator? Somebody
must be thinking similar. It seems like it would be easier to
implement the () operator than handle procs/methods specially.
And cleaner.

> At the very least, no one has been able to explain to me why
> what
> Ruby does isn't acceptable. I don't want or need Ruby to be
> Python
> and I don't want or need a () operator.

I didn't know Python had that. Just found it:

x(arg1, arg2, ...) is a shorthand for x.__call__(arg1, arg2,
...)





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1 Answer

gabriele renzi

5/19/2005 9:14:00 AM

0

Eric Mahurin ha scritto:

>
> I didn't know Python had that. Just found it:
>
> x(arg1, arg2, ...) is a shorthand for x.__call__(arg1, arg2,
> ..)

but notice that python does not have (and never had) separate namespaces
for methods and variables, like we have in ruby.
Special casing #call (which is something I like) will cause some
breakage in cases where you do things like
meth = meth(meth)
or similar, see a recent message from Joao Pedrosa (IIRC) that shows
some examples of real libraries getting an error.