--- Additional Comments From kreckel at ginac dot de 2005-04-18 21:08
---
(In reply to comment #5)
> (In reply to comment #3)
> > This sentence just says that you can't do this:
> > class A { private: struct I{}; };
> > class B { friend class A::I; };
> > because A::I isn't acce
--- Additional Comments From kreckel at ginac dot de 2005-04-18 21:04
---
(In reply to comment #3)
> This sentence just says that you can't do this:
> class A { private: struct I{}; };
> class B { friend class A::I; };
> because A::I isn't accessible in B.
Where's that snippet
--- Additional Comments From bangerth at dealii dot org 2005-04-18 20:34
---
And the answer to the original question: even though gcc handled it as this
in the past (incorrectly), a friend declaration is not a declaration. In
other words, if a name is not declared before, but first see
--- Additional Comments From bangerth at dealii dot org 2005-04-18 20:29
---
This sentence just says that you can't do this:
class A { private: struct I{}; };
class B { friend class A::I; };
because A::I isn't accessible in B.
W.
--
http://gcc.gnu.org/bugzilla/show_bug.c
--- Additional Comments From kreckel at ginac dot de 2005-04-18 20:19
---
(In reply to comment #1)
> No, the code is invalid, as Y has not been interjected yet. This is a
progression and not a regression.
Really? What about paragraph 11.4/7 "A name nominated by a friend
declaration sh
--- Additional Comments From pinskia at gcc dot gnu dot org 2005-04-18
19:40 ---
No, the code is invalid, as Y has not been interjected yet. This is a
progression and not a regression.
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