🚀 go-pugleaf

kanze@gabi-soft.fr (J. Kanze) writes:


>Matt Austern (matt@godzilla.EECS.Berkeley.EDU) wrote:

>|> In article <4638jp$ef4@fsgm01.fnal.gov> b91926@fsgm01.fnal.gov (David Sachs) writes:

>|> > X x(a) uses higher priority conversion by overloading rules
>|> > 
>|> > X x=a uses X(a.operator X&)

>|> Sort of.  If a is of class A, if the copy constructor X::X(const X&)
>|> exists, and if the converting constructor X::X(const A&) exists and
>|> isn't qualified as explicit, then "X x(a)" simply invokes the
>|> converting constructor while "X x=a" invokes the converting
>|> constructor and then the copy constructor.

>Except that there is no guarantee that the copy constructor will
>actually be invoked.  The compiler is allowed to optimize it away.

>|> Section 12.6 of the working paper explains this in detail.  The rest
>|> of chapter 12 is somewhat relevant too.

>Concerning the original posting: I don't think it will work.  As I
>understand what you are suggesting:

>    class B {} ;
>    class D : public B { operator B() ; } ;

>    D               aD ;
>    B               aB( aD ) ;
>    B               anotherB = aD ;

>What you are saying is that in the first case, D::operator B() will not
>be considered, and in the second case, it will be used.

>In fact, in both cases, it will be considered, but as a user defined
>conversion, it is pretty far down on the scale; in both cases, the
>derived->base conversion will have precedence.  (I believe.)
>  is summarized in http://dogbert.lbl.gov/~matt/std-c++/policy.html. ]

I am relying on paragraph 2 of scetion 12.3.2 of the proposed standard,
which appears to specify that a conversion function will be used in
certain cases. Possibly, the draft standard is ambiguous in this area.
-- 
** The Klingons' favorite food was named by the first earthling to see it **
David Sachs - Fermilab, HPPC MS369 - P. O. Box 500 - Batavia, IL 60510
Voice: 1 708 840 3942      Deparment Fax: 1 708 840 3785


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