------- Additional Comments From giovannibajo at libero dot it 2005-01-06
10:13 -------
This is invalid. Within the expression "C1<T>::nested", T cannot be deduced
because it is in an nondeduced context.
ISO/ANSI C++ standard reference: [temp.deduct.type]/4:
The nondeduced contexts are:
-- The nested-name-specifier of a type that was specified using a
qualified-id.
-- A type that is a template-id in which one or more of the template-
arguments is an expression that references a template-parameter.
When a type name is specified in a way that includes a nondeduced
context, all of the types that comprise that type name are also
nondeduced. However, a compound type can include both deduced and
nondeduced types. [Example: If a type is specified as A<T>::B<T2>,
both T and T2 are nondeduced. Likewise, if a type is specified as
A<I+J>::X<T>, I, J, and T are nondeduced. If a type is specified as
void f(typename A<T>::B, A<T>), the T in A<T>::B is nondeduced but the
T in A<T> is deduced.
Also notice the end of paragraph 3: "If a template parameter is used only in
nondeduced contexts and is not explicitly specified, template argument
deduction fails."
Using a specialization as a workaround is a bit bloating. The correct way is to
explicitly specify the template arguments which cannot be deduced when calling
the template. In your case, you can use f2<float>(n).
--
What |Removed |Added
----------------------------------------------------------------------------
Status|UNCONFIRMED |RESOLVED
Resolution| |INVALID
http://gcc.gnu.org/bugzilla/show_bug.cgi?id=19288
