static member object initialization - c++

Barry wrote:

> Comeau online also accepts all the cases.
> Waiting guru for explanation.

Hey, no guru, but consider this little piece that
also compiles (and should)
:
class A
{
static A a;
A() {}
};

A A::a; //The same as A A::a = A();
int main(){}

The code above calls a constructor that's also
private. Why should it not be able to call a
static member function for the same reason?

I suppose c++98, Par 3.4.1:12 could be it:

A name used in the definition of a static data member
of class X (...) is looked up as if the name was used in
a member function of X. Boy did they think of <almost>
everything :-).

Regards,

Werner







>
>
> --
> Thanks
> Barry

werasm wrote:
> Barry wrote:
>
>> Comeau online also accepts all the cases.
>> Waiting guru for explanation.
>
> Hey, no guru, but consider this little piece that
> also compiles (and should)
> :
> class A
> {
> static A a;
> A() {}
> };
>
> A A::a; //The same as A A::a = A();
> int main(){}
>
> The code above calls a constructor that's also
> private. Why should it not be able to call a
> static member function for the same reason?
>
> I suppose c++98, Par 3.4.1:12 could be it:
>
> A name used in the definition of a static data member
> of class X (...) is looked up as if the name was used in
> a member function of X. Boy did they think of <almost>
> everything :-).
>

Well, no need to compare with Bjarne, or ... to be guru. :-)
Actually, I shouldn't post the word, it's like "an very simple question"
-- it hurts if the readers don't even know that issue.Putting "guru
help" is likely to drive some people away
:-)


--
Thanks
Barry

On Sep 25, 1:47 pm, "subramanian10...@yahoo.com, India"
<subramanian10...@yahoo.com> wrote:
> Consider the following program:

> #include <iostream>

> using namespace std;

> class Test
> {
> static Test t;
> static Test init_Test( ) { return t; }
> Test(const Test & rhs) { cout << "copy ctor" << endl; }
> };

> Test Test::t = init_Test( );

> int main()
> {
> return 0;
> }

> This program compiles fine under both g++ and VC++2005 Express
> Edition and both produce the following same output
> copy ctor

Which doesn't really surprise me, knowing roughly what the
compilers will generate.

> However consider the statement
> Test Test::t = init_Test( );
> Here init_Test( ) is called which returns the static member object
> under construction which is 't'. I do not understand how we can
> return an object which is still under construction.

You can't. It's undefined behavior.

> How is it accepted by the compiler ?

It's accepted by the compiler because the compiler is not
required to diagnose the error. It's accepted, in fact, because
the C++ is designed to allow separate compilation. There's
nothing in the line you cite which causes problems per se, given
that the compiler doesn't know what's in init_Test at that
point. (init_Test could return a local static, or even a newly
constructed object each time.) And there's nothing which causes
a problem per se in init_Test, since most uses of it would be
OK. It's only the combination of the two (this particular
implementation of init_Test(), used in this particular context)
which causes problems, and the language specification is
carefully designed so that the compiler is never required to
look into a function which is being called.

In practice, you're code works because the copy constructor
doesn't actually use the object it's copying. If it did, it
would find it uninitialized (or rather, zero initialized, since
this is a static object).

--
James Kanze (GABI Software) email:james.kanze@gmail.com
Conseils en informatique orientée objet/
Beratung in objektorientierter Datenverarbeitung
9 place Sémard, 78210 St.-Cyr-l'École, France, +33 (0)1 30 23 00 34

On Sep 25, 3:36 pm, Jonathan Lane <jonathan.la...@googlemail.com>
wrote:
> On Sep 25, 2:02 pm, Barry <dhb2...@gmail.com> wrote:
> > Victor Bazarov wrote:
> > > subramanian10...@yahoo.com wrote:
> > >> Consider the following program:

> > >> #include <iostream>

> > >> using namespace std;

> > >> class Test
> > >> {
> > >> static Test t;
> > >> static Test init_Test( ) { return t; }
> > >> Test(const Test & rhs) { cout << "copy ctor" << endl; }
> > >> };

> > >> Test Test::t = init_Test( );

> > >> int main()
> > >> {
> > >> return 0;
> > >> }

> > >> This program compiles fine under both g++ and VC++2005 Express Edition
> > >> and both produce the following same output
> > >> copy ctor

> > >> However consider the statement
> > >> Test Test::t = init_Test( );

> I would think that:
> Test Test::t <- this part runs the default ctor
> = init_Test() <- this part then calls operator=() on the newly
> constructed Test object.

You would think wrong. There's no assignment here. We have a
case of a punctuation symbol (the '=') which has two very
different meanings. In an expression, it means assignment, but
the expression here doesn't begin until after the '=' sign, and
in a declaration, the '=' sign is used to specify copy
initialization (which is the same as value initialization when
the initialization expression has the same type as the object
being initialized). The above statement has exactly the same
meaning as:

Test Test::t( init_Test() ) ;

> init_Test then returns t by value so you end
> up with a statement like:
> Test Test::t = Test(Test::t);
> Or am I missing the point here?

I think you've misunderstood the C++ definition syntax.

--
James Kanze (GABI Software) email:james.kanze@gmail.com
Conseils en informatique orientée objet/
Beratung in objektorientierter Datenverarbeitung
9 place Sémard, 78210 St.-Cyr-l'École, France, +33 (0)1 30 23 00 34

On Sep 25, 3:39 pm, Barry <dhb2...@gmail.com> wrote:
[...]
> My point is that whichever "Init()" or "A::Init()" is private
> member function, why access control does not apply here.

Access control applies, but you're initializing a member
variable, so you are logically within the class.

--
James Kanze (GABI Software) email:james.kanze@gmail.com
Conseils en informatique orientée objet/
Beratung in objektorientierter Datenverarbeitung
9 place Sémard, 78210 St.-Cyr-l'École, France, +33 (0)1 30 23 00 34

On Sep 26, 9:09 am, James Kanze <james.ka...@gmail.com> wrote:
> On Sep 25, 3:36 pm, Jonathan Lane <jonathan.la...@googlemail.com>
> wrote:
>
>
>
> > On Sep 25, 2:02 pm, Barry <dhb2...@gmail.com> wrote:
> > > Victor Bazarov wrote:
> > > > subramanian10...@yahoo.com wrote:
> > > >> Consider the following program:
> > > >> #include <iostream>
> > > >> using namespace std;
> > > >> class Test
> > > >> {
> > > >> static Test t;
> > > >> static Test init_Test( ) { return t; }
> > > >> Test(const Test & rhs) { cout << "copy ctor" << endl; }
> > > >> };
> > > >> Test Test::t = init_Test( );
> > > >> int main()
> > > >> {
> > > >> return 0;
> > > >> }
> > > >> This program compiles fine under both g++ and VC++2005 Express Edition
> > > >> and both produce the following same output
> > > >> copy ctor
> > > >> However consider the statement
> > > >> Test Test::t = init_Test( );
> > I would think that:
> > Test Test::t <- this part runs the default ctor
> > = init_Test() <- this part then calls operator=() on the newly
> > constructed Test object.
>
> You would think wrong. There's no assignment here. We have a
> case of a punctuation symbol (the '=') which has two very
> different meanings. In an expression, it means assignment, but
> the expression here doesn't begin until after the '=' sign, and
> in a declaration, the '=' sign is used to specify copy
> initialization (which is the same as value initialization when
> the initialization expression has the same type as the object
> being initialized). The above statement has exactly the same
> meaning as:
>
> Test Test::t( init_Test() ) ;
>
> > init_Test then returns t by value so you end
> > up with a statement like:
> > Test Test::t = Test(Test::t);
> > Or am I missing the point here?
>
> I think you've misunderstood the C++ definition syntax.
Indeed I did. I realized my error last night. I think I missed the
point that it was a static and the statement was outside of the main
in this case. Anyway, I see the point now. Thanks.