sergey@cis.ohio-state.edu (Sergey Zhupanov) wrote:
> Folks,
>
> In the example in Draft Standard Dec. 2, 1996, page 3-19, the example
> on top of the page:
>
> The line that reads:
>         e.B::a = 0;       // ok, only one A:a in E
>
> since B is defined as:
>         struct B: virtual A { };
>
> isn't A::a private in B, and therefore inaccessible in e?

No. Public is assumed when the derived class is a struct.
Private is the default for a class.

Paul
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Sergey Zhupanov (sergey@cis.ohio-state.edu) wrote:
: Folks,

: In the example in Draft Standard Dec. 2, 1996, page 3-19, the example
: on top of the page:

: The line that reads:
:  e.B::a = 0;       // ok, only one A:a in E

: since B is defined as:
:  struct B: virtual A { };

: isn't A::a private in B, and therefore inaccessible in e?

No.  A is a struct, so it's default access is public [class.access],
and B is a struct so by default it's base classes access is public
[class.access.base]

--
        Tony Cook - tony@online.tmx.com.au
                    100237.3425@compuserve.com
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Folks,

In the example in Draft Standard Dec. 2, 1996, page 3-19, the example
on top of the page:

The line that reads:
 e.B::a = 0;       // ok, only one A:a in E

since B is defined as:
 struct B: virtual A { };

isn't A::a private in B, and therefore inaccessible in e?

thanks,
  sergey
Thanks,
  sergey
----------------------------------
Email: sergey@cis.ohio-state.edu
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