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[Bug c++/46690] New: Using declaration of a dependent name
- From: "ilpoilves at hotmail dot com" <gcc-bugzilla at gcc dot gnu dot org>
- To: gcc-bugs at gcc dot gnu dot org
- Date: Sun, 28 Nov 2010 09:35:47 +0000
- Subject: [Bug c++/46690] New: Using declaration of a dependent name
- Auto-submitted: auto-generated
http://gcc.gnu.org/bugzilla/show_bug.cgi?id=46690
Summary: Using declaration of a dependent name
Product: gcc
Version: 4.4.5
Status: UNCONFIRMED
Severity: normal
Priority: P3
Component: c++
AssignedTo: unassigned@gcc.gnu.org
ReportedBy: ilpoilves@hotmail.com
Consider the following code:
template <typename T>
class A {public: typedef int C;};
template <typename T>
class B: public A<T>
{
public:
using typename A<T>::C;
C operator[](int i) const { return 0; }
};
int main()
{
B<int> b;
b[2];
return 0;
}
The compilation gives:
test.cpp:9: error: ISO C++ forbids declaration of 'C' with no type
test.cpp:9: error: expected ';' before 'operator'
test.cpp:10: error: expected ';' before '}' token
test.cpp: In function 'int main()':
test.cpp:15: error: no match for 'operator[]' in 'b[2]'
A discussion in comp.lang.c++ revealed that this feature was underspecified in
C++03. I propose the using declaration either just to work, or report a proper
error.
To see the intent, consider the current wording for C++0x:
From n3126 (C++0x draft from August 2010), Â7.3.3:
"20. If a using-declaration uses the keyword typename and specifies a
dependent name (14.6.2), the name introduced
by the using-declaration is treated as a typedef-name (7.1.3)."
So the work-around here is
typedef typename A<T>::C C;
which is equivalent.
The code above is accepted at least by Visual Studio 2008, 2010, and Comeau
C++, all in strict mode.