Re: Inline destructors and construtors

From:
James Kanze <james.kanze@gmail.com>
Newsgroups:
comp.lang.c++
Date:
Fri, 30 Jan 2009 15:25:51 -0800 (PST)
Message-ID:
<1b39557e-0ccb-4ab7-9ec3-69d3ffc35bde@z28g2000prd.googlegroups.com>
On 30 jan, 12:28, "Alf P. Steinbach" <al...@start.no> wrote:

* James Kanze:

I'm not sure what "to be called virtually" means. I've
never seen the expression in any serious technical
literature


Ah, terminology!

Always fun!


Isn't it:-).

Well, the *international C++ standard* ISO/IEC 14882 uses the
term "virtual call" in two places.


Interesting. Does it define it, or just use it? (If the
latter, I would consider this a defect.)

The C++ standard is, IMHO, "serious technical literature"... ;-)


Obviously, I haven't memorized it completely:-). The full
explination of what happens when you call a function, if the
function happens to be virtual (in the static type) speaks of
resolution against the dynamic type, which is fairly precise.
For some strange reason, I supposed that they would be
consistent in their use of terminology:-).

The original Smalltalk view is very clear on the effect of a
virtual call. When there is a source code call of a method
with signature M on an object O, and the method M is virtual
in the statically known type of O, the execution effect is as
if a search for M starts in the most derived class of O's
dynamic type. The search for M goes upward all available base
class chains, in any order, until an M definition is found
(the order doesn't matter because C++ guarantees that there
will be only one M reachable in this way, although higher up
there may be more). The found M is called with a pointer to O
as the 'this' pointer.


I'm a bit confused about this. First you speak of Smalltalk,
then C++. In Smalltalk, IIRC, there can't be any ambiguity
because Smalltalk doesn't support multiple inheritance. And
IIRC (admittedly, a very big if), all functions in Smalltalk are
virtual. (Actually, IIRC, Smalltalk doesn't have functions,
only methods. And you can't call a method; you send a message
to an object. But I've never actually used Smalltalk, so none
of this is fixed in my mind like it is for C++.)

PS: Yeah, I know you know what a virtual call is, even if you
disagree with the standard's terminology (as do I for some
other terms). The explanation above is for the benefit of
other readers -- not all are aware of the Smalltalk view.


To tell the truth, I'm not that familiar with it either.

--
James Kanze

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