[Bug c++/32039] Using declaration accepts non-visible members from base classes
https://gcc.gnu.org/bugzilla/show_bug.cgi?id=32039 Richard Biener changed: What|Removed |Added Status|NEW |ASSIGNED
[Bug c++/32039] Using declaration accepts non-visible members from base classes
https://gcc.gnu.org/bugzilla/show_bug.cgi?id=32039 Ville Voutilainen ville.voutilainen at gmail dot com changed: What|Removed |Added Status|ASSIGNED|RESOLVED CC||ville.voutilainen at gmail dot com Resolution|--- |INVALID --- Comment #7 from Ville Voutilainen ville.voutilainen at gmail dot com --- This is invalid. [namespace.udecl]/3 says that if the declaration names a constructor, the nested-name-specifier shall name a direct base, but if the declaration names something else than a constructor, indirect bases are fine. The name is looked up by member name lookup, which will look in A first, and the hiding in B does not matter. Clang agrees with this interpretation.
[Bug c++/32039] Using declaration accepts non-visible members from base classes
https://gcc.gnu.org/bugzilla/show_bug.cgi?id=32039 Ville Voutilainen ville.voutilainen at gmail dot com changed: What|Removed |Added Status|RESOLVED|NEW Resolution|INVALID |--- --- Comment #9 from Ville Voutilainen ville.voutilainen at gmail dot com --- Pardon that, I failed to notice that part (and yes, sorry I missed it in the original report). Back to open, then. I think the standard needs to be clarified about what it wants. :)
[Bug c++/32039] Using declaration accepts non-visible members from base classes
https://gcc.gnu.org/bugzilla/show_bug.cgi?id=32039 --- Comment #8 from Harald van Dijk harald at gigawatt dot nl --- (In reply to Ville Voutilainen from comment #7) This is invalid. [namespace.udecl]/3 says that if the declaration names a constructor, the nested-name-specifier shall name a direct base, but if the declaration names something else than a constructor, indirect bases are fine. Yes, but you're ignoring p14 (now p17 in N4140) which was mentioned right in the initial report, which adds The base class members mentioned by a using-declaration shall be visible in the scope of at least one of the direct base classes of the class where the using-declaration is specified. That is not limited to constructors, that is a separate requirement in the standard that GCC and clang both fail to implement.
[Bug c++/32039] Using declaration accepts non-visible members from base classes
https://gcc.gnu.org/bugzilla/show_bug.cgi?id=32039 --- Comment #10 from Ville Voutilainen ville.voutilainen at gmail dot com --- Ok, this is http://open-std.org/JTC1/SC22/WG21/docs/cwg_closed.html#1960 which says The rule was introduced because the hiding of a base class member by an intermediate derived class is potentially intentional and should not be capable of circumvention by a using-declaration in a derived class. The consensus of CWG preferred not to change the restriction. So yes, we need to fix this. :)
[Bug c++/32039] Using declaration accepts non-visible members from base classes
http://gcc.gnu.org/bugzilla/show_bug.cgi?id=32039 --- Comment #2 from fabien at gcc dot gnu.org --- (In reply to Andrew Stubbs from comment #0) The problem should be that B::foo hides A::foo from class C. Clause 7.3.3/14 of the C++ standard says the using declaration should not work, in this case - class A is not a direct base class of class C. However, GCC 4.1.1 accepts it with no diagnostic. 7.3.3/14 says ...The base class members mentioned by a using-declaration shall be visible in the scope of at least one of the direct base classes of the class where the using-declaration is specified... In the example above, A (from using A::foo) is visible from its direct base class B. Consequently, it is valid. Adding 'using A::foo' within B does not change anything to that. without 'using A::foo': 'int B::foo(long)' hides int A::foo(int) in B and 'using A::foo' brings 'int A::foo(int)' into C and hides 'int B::foo(long)'. with 'using A::foo': two overloads of 'foo' are present in B: 'int B::foo(long)' and 'int A::foo(int)' (brought into B scope by the using-declaration). and 'using A::foo' brings 'int A::foo(int)' into C and hides 'int B::foo(long)' (and 'using A::foo)
[Bug c++/32039] Using declaration accepts non-visible members from base classes
http://gcc.gnu.org/bugzilla/show_bug.cgi?id=32039 --- Comment #3 from fabien at gcc dot gnu.org --- (In reply to Eelis from comment #1) Still accepted by 4.4. Comeau concurs with reporter, and rejects saying: line 15: error: class member designated by a using-declaration must be visible in a direct base class Which seems wrong according to the standard quoted above. From Clang results and the analysis done in bug 19377, I am inclined to close this bug as invalid.
[Bug c++/32039] Using declaration accepts non-visible members from base classes
http://gcc.gnu.org/bugzilla/show_bug.cgi?id=32039 fabien at gcc dot gnu.org changed: What|Removed |Added Status|ASSIGNED|RESOLVED Resolution|--- |INVALID --- Comment #4 from fabien at gcc dot gnu.org --- Closed as invalid.
[Bug c++/32039] Using declaration accepts non-visible members from base classes
http://gcc.gnu.org/bugzilla/show_bug.cgi?id=32039 Harald van Dijk harald at gigawatt dot nl changed: What|Removed |Added CC||harald at gigawatt dot nl --- Comment #5 from Harald van Dijk harald at gigawatt dot nl --- This bug is about visibility, bug 19337 is about accessibility. I incorrectly used visibility in my comment on the other bug, I apologise if that has confused matters. The comments there do not apply here. Normally, a qualified name such as A::foo can be used to refer to a hidden member. In this instance, however, the standard makes a special exception, and states that A::foo must be visible in B as well, even though a qualified name is used, for the using declaration to be valid. In B, int A::foo(int) is hidden by int B::foo(long) because B's member function has the same name. (3.3.10p1) A name is said to be visible, if it is in scope, and not hidden. (3.3.10p5) So I think this bug report is valid and unrelated to 19337: A's member function is indeed not visible in any of the direct bases of C. Your comment (the In the example above, A (from using A::foo) is visible bit) suggests that you read the standard as requiring that the base class be visible, rather than the base class member.
[Bug c++/32039] Using declaration accepts non-visible members from base classes
http://gcc.gnu.org/bugzilla/show_bug.cgi?id=32039 fabien at gcc dot gnu.org changed: What|Removed |Added Status|RESOLVED|ASSIGNED Resolution|INVALID |--- --- Comment #6 from fabien at gcc dot gnu.org --- (In reply to Harald van Dijk from comment #5) [...] Your comment (the In the example above, A (from using A::foo) is visible bit) suggests that you read the standard as requiring that the base class be visible, rather than the base class member. Obviously should read members of A are visible... Anyway thanks for the clarification.
[Bug c++/32039] Using declaration accepts non-visible members from base classes
http://gcc.gnu.org/bugzilla/show_bug.cgi?id=32039 fabien at gcc dot gnu.org changed: What|Removed |Added Status|UNCONFIRMED |ASSIGNED Last reconfirmed||2011-11-22 Ever Confirmed|0 |1
[Bug c++/32039] Using declaration accepts non-visible members from base classes
--- Comment #1 from gcc-bugzilla at contacts dot eelis dot net 2008-08-21 12:00 --- Still accepted by 4.4. Comeau concurs with reporter, and rejects saying: line 15: error: class member designated by a using-declaration must be visible in a direct base class -- gcc-bugzilla at contacts dot eelis dot net changed: What|Removed |Added CC||gcc-bugzilla at contacts dot ||eelis dot net http://gcc.gnu.org/bugzilla/show_bug.cgi?id=32039
