http://gcc.gnu.org/bugzilla/show_bug.cgi?id=51336
Daniel Krügler <daniel.kruegler at googlemail dot com> changed:
What |Removed |Added
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--- Comment #1 from Daniel Krügler <daniel.kruegler at googlemail dot com>
2011-11-28 20:49:52 UTC ---
(In reply to comment #0)
> #include <type_traits>
> template<class T>
> struct A {
> template<class=typename
> std::enable_if<std::is_same<T,int>::value>::type>
> A(A const&){}
> };
> constexpr bool b = std::is_abstract<A<double>>::value;
I *think* the compiler is right to reject this as it currently does, we have
*no* sfinae here. When you instantiate A<double>, the declaration of the
template constructor is also instantiated, but at that point A<double> is an
incomplete type.
IMO you need one further indirection, e.g.
template<class T>
struct A {
template<class U = T, class = typename
std::enable_if<std::is_same<U, int>::value>::type
>
A(A const&){}
};
Btw.: Neither of these forms can ever prevent the "real" copy constructor to be
declared, defined, and used by the compiler.
> I am not sure what is supposed to happen (that's why I tried), but this result
> doesn't seem right. Filed under C++ because is_abstract directly forwards to
> the __is_abstract builtin, but feel free to reassign to libstdc++ if you think
> the problem is there somehow.
Lets look what the compiler-intrinsic people say, above is my first guess on
that.
