Francis Glassborow wrote:
>
> In article <383BE366.65BC@worldnet.att.net>, Tamasa Mahapatra <saroj-
> tamasa@worldnet.att.net> writes
> >Hi,
> >
> >If I have a construct like
> >
> >C :: b and C is a class and b is an object, how do I know if
> >the scope resolution operator applies to C (member b in class
> >or namespace C) or b (object in global scope of type C or start
> >of a function b returning C)?
>
> Because the syntax does not allow you to declare global objects that
> way.

That's not quite correct. There is an ambiguity in the syntax, and
the meaning of
 C :: b;
is not well-defined. It is best if you don't need a global qualifier
on the declarator. If you do, use parens to disambiguate:
 C (::b); // declares global b of type C
 C::b (::a); // declares global a of type C::b
--
Steve Clamage, stephen.clamage@sun.com
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Hi,

If I have a construct like

C :: b and C is a class and b is an object, how do I know if
the scope resolution operator applies to C (member b in class
or namespace C) or b (object in global scope of type C or start
of a function b returning C)?

Thanks,
Saroj
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In article <383BE366.65BC@worldnet.att.net>, Tamasa Mahapatra <saroj-
tamasa@worldnet.att.net> writes
>Hi,
>
>If I have a construct like
>
>C :: b and C is a class and b is an object, how do I know if
>the scope resolution operator applies to C (member b in class
>or namespace C) or b (object in global scope of type C or start
>of a function b returning C)?

Because the syntax does not allow you to declare global objects that
way.


Francis Glassborow      Journal Editor, Association of C & C++ Users
64 Southfield Rd
Oxford OX4 1PA          +44(0)1865 246490
All opinions are mine and do not represent those of any organisation


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