While coding around in a project I'm working on, I discovered something really odd:
namespace detail {
struct tuplelike_tag { };
struct arraylike_tag { };
template<typename>
struct call_with_traits;
template<typename... Ts>
struct call_with_traits<std::tuple<Ts...>> {
using tag = tuplelike_tag;
enum { size = sizeof...(Ts) };
};
template<typename T, std::size_t Sz>
struct call_with_traits<std::array<T, Sz>> {
using tag = arraylike_tag;
enum { size = Sz };
};
template<typename T, std::size_t Sz>
struct call_with_traits<T[Sz]> {
using tag = arraylike_tag;
enum { size = Sz };
};
template<typename F, typename T, int... Is>
auto call_with(F && f, T && tup, indices<Is...>, tuplelike_tag) -> ResultOf<Unqualified<F>> {
return (std::forward<F>(f))(std::get<Is>(std::forward<T>(tup))...);
}
template<typename F, typename A, int... Is>
auto call_with(F && f, A && arr, indices<Is...>, arraylike_tag) -> ResultOf<Unqualified<F>> {
return (std::forward<F>(f))(std::forward<A>(arr)[Is]...);
}
}
template<typename F, typename Cont>
inline auto call_with(F && f, Cont && cont) -> ResultOf<Unqualified<F>> {
using unqualified = Unqualified<Cont>;
using traits = typename detail::call_with_traits<unqualified>;
using tag = typename detail::call_with_traits<unqualified>::tag;
using no_tag = typename traits::tag; // this is what it's all about
return detail::call_with(std::forward<F>(f), std::forward<Cont>(cont), build_indices<traits::size>(), tag());
}
The odd thing about this code is that tag gets resolved to typename detail::call_with_traits::tag;, but no_tag errors out:
error: no type named ‘tag’ in ‘using traits = struct detail::call_with_traits<typename std::remove_cv<typename std::remove_reference<_To>::type>::type>’
even though it should refer to the same using declaration in the same struct. Is there something I'm missing, or is this some bug in GCC?
A live example can be found here, including the related error message from GCC.
This appears to be a bug in g++ 4.7.2. Here is a minimal example:
template<typename> struct A { using tag = int; };
template<typename T>
inline void f() {
using AT = A<T>;
typename A<T>::tag x; // no error
typename AT::tag y; // error
}
int main(int argc, char ** argv) {
f<int>();
return 0;
}
There are no errors if a typedef
is used instead of a using
declaration, or if you use A<int>
instead of A<T>
.
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With