Re: Ripple effect

[Robert A Duff <bobduff@shell01.TheWorld.com>]

"Jeffrey R. Carter" <spam.not.jrcarter@acm.not.spam.org> writes:

> Robert A Duff wrote:
> > The "Nada" thing was a joke.  I would not call a language "Nada", and
> > certainly not "Duff".  ;-)
> 
> Nada sounds like a great name for a language :) "What's so great about
> this language?" "Oh, Nada."

Nah.

> You did indicate that you would probably name it after someone.

Who or whom?  ;-)

> > I think you mean there is no equivalent of "use type".
> 
> IIRC, there is no equivalent of "use" or "use type". Of course, I may
> have misunderstood.

Not sure about "use" but "use type" is a bad idea -- should be the
default.

> > No, I think you misunderstand the "Ripple Effect".  As I understand it,
> > the Ripple Effect means that adding/removing a with_clause can cause
> > compilation units that do not depend DIRECTLY on the modified thing to
> > become illegal.  For example, suppose C with's B and B with's A.  Can a
> > with_clause on A affect the legality of C?  If so, there's a Ripple
> > Effect.
> 
> I've just rechecked the answer in the FAQ at adapower.com, and I think
> we're both mistaken: 

Could be.

>...'In brief, the (undesirable) Ripple effect was
> related to whether the legality of a compilation unit could be affected
> by adding or removing an otherwise unneeded "with" clause on some
> compilation unit on which the unit depended, directly or indirectly.'
> (Tucker Taft)

Sorry, but I've lost the context here.  What does "otherwise unneeded"
mean?

> So it's not a with on C, as I thought, and can be a with on B, which you
> exclude. It also refers only to unneeded withs. So, if B withs A
> unnecessarily, that could cause a Ripple effect.

Not sure what "unnecessarily" means.  I mean, if you say "with A" and
refer to A.Mumble, that's "necessary", right?

> > Anyway, to answer Jeff's question:  I think with_clauses should be
> > transitive in the first place, so that the Ripple Effect is not an
> > issue.
> 
> So you mean, in the example you give above, that because C withs B and B
> withs A, that C withs A? Is that how your language works?

Yes.  Do you think it should be otherwise?

- Bob