Joseph Heled <joseph@itgssi.com> wrote in
news:3CF277C1.B98D3211@itgssi.com:

>
> Is this legal or not? g++ 3.1 says not, but I can't see why.
> I would appreciate a comment from a "language lawyer" ...
>
> ----------------------------------------------
> class A {
> private:
>   class B {
>     friend void f(B&);
>   };
> };
>
> void
> f(A::B&) {}
> ----------------------------------------------
>
> =================================================
> g++ -Wall -c t3.cc
> t3.cc: In function `void f(A::B&)':
> t3.cc:4: `class A::B' is private
> t3.cc:11: within this context
> =================================================
>

Both your classes is legal code. Since B is private, only A is able to create
an object of B and thus is the only one able to call 'f(A::B&)'. It looks
like you're calling 'f' from the global scope in your example - how did it
get access to a 'B'-object?

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On Mon, 27 May 2002, Joseph Heled wrote:
>
> class A {
> private:
>   class B {
>     friend void f(B&);
>   };
> };
>
> void
> f(A::B&) {}
>
> =================================================
> g++ -Wall -c t3.cc
> t3.cc: In function `void f(A::B&)':
> t3.cc:4: `class A::B' is private
> t3.cc:11: within this context
> =================================================
>
> Artur Siekielski claimed,
>
>  This declaration says, that 'void f(B&)' has access to private and protected
>  members of 'B' class, it says anything about accessing objects of type 'B'.
>
> And suggested I would repeat the friendship relation inside A.

That is correct, under current language rules.

But a proposal is before the C++ Committee that would give member
classes full access to the class itself. Under that modifed language
rule, the code would be valid. I believe the proposal will be adopted
in a Technical Corrigendum to the standard, but it will not be in the
TC which is currently in progress.

--
Steve Clamage, stephen.clamage@sun.com

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Is this legal or not? g++ 3.1 says not, but I can't see why.
I would appreciate a comment from a "language lawyer" ...

----------------------------------------------
class A {
private:
  class B {
    friend void f(B&);
  };
};

void
f(A::B&) {}
----------------------------------------------

=================================================
g++ -Wall -c t3.cc
t3.cc: In function `void f(A::B&)':
t3.cc:4: `class A::B' is private
t3.cc:11: within this context
=================================================

Artur Siekielski claimed,

 This declaration says, that 'void f(B&)' has access to private and protected
 members of 'B' class, it says anything about accessing objects of type 'B'.

And suggested I would repeat the friendship relation inside A.

I would think it ironic if 'f' can access protected/private parts of B, but
can't access a B itself.
Of course I was trying to define operator<< (i.e. 'friend operator<<(ostream&,
B&)'), and worse, his suggestion fails in my case, since I am using forwarded
nested classes.


class A {
private:
  class B;
  // *CAN'T* put  'friend void f(A::B::C&);' here, since B body is not know yet.
};

class A::B {
private:
  class C;
  // I can put 'friend void f(C&);' here
};

class A::B::C {
public:
  friend void f(C&);
};


I think my example shows that if this is the standard, perhaps it is a slightly
broken? I would expect the granting of friendship to automatically grant access
to the class itself.


Thanks. Joseph

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