constexprテンプレート引数奇妙さ

Question

GCC（5.3）& Clang（3.8）はtestの最初の行が悪いと主張しますが、2番目の行は問題ありません。 MSVC（2015.2）によれば、両方とも無効です。

template< typename N, typename T > 
void f(N n, T t) { std::get<n>(t); } 
void test() { 
    std::get< std::integral_constant< size_t, 0 >() >(std::make_tuple(123)); // not ok 
    f(std::integral_constant< size_t, 0 >(), std::make_tuple(123)); // ok for gcc, clang, but not msvc 
} 

正確には、標準によると違いはありますか？このコードは最初から正当なものですか？最初の行のための

---

打ち鳴らすエラー：

In file included from main.cpp:2: 
/usr/local/bin/../lib/gcc/x86_64-unknown-linux-gnu/5.3.0/../../../../include/c++/5.3.0/tuple:874:34: error: no matching function for call to '__get_helper2' 
    { return std::forward<_Tp&&>(std::__get_helper2<_Tp>(__t)); } 
           ^~~~~~~~~~~~~~~~~~~~~~~ 
main.cpp:10:10: note: in instantiation of function template specialization 'std::get<std::integral_constant<unsigned long, 0>(), int>' requested here 
    std::get<std::integral_constant<size_t, 0>()>(std::make_tuple(123)); // not ok 
     ^
/usr/local/bin/../lib/gcc/x86_64-unknown-linux-gnu/5.3.0/../../../../include/c++/5.3.0/tuple:856:5: note: candidate template ignored: could not match '_Tuple_impl' against 'tuple' 
    __get_helper2(_Tuple_impl<__i, _Head, _Tail...>& __t) noexcept 
    ^
/usr/local/bin/../lib/gcc/x86_64-unknown-linux-gnu/5.3.0/../../../../include/c++/5.3.0/tuple:861:5: note: candidate template ignored: could not match '_Tuple_impl' against 'tuple' 
    __get_helper2(const _Tuple_impl<__i, _Head, _Tail...>& __t) noexcept 
    ^
1 error generated. 

gccのエラー：

In file included from main.cpp:2:0: 
/usr/local/include/c++/5.3.0/tuple: In instantiation of 'constexpr _Tp&& std::get(std::tuple<_Elements ...>&&) [with _Tp = std::integral_constant<long unsigned int, 0ul>(); _Types = {int}]': 
main.cpp:10:75: required from here 
/usr/local/include/c++/5.3.0/tuple:874:57: error: no matching function for call to '__get_helper2(std::tuple<int>&)' 
    { return std::forward<_Tp&&>(std::__get_helper2<_Tp>(__t)); } 
                 ^
/usr/local/include/c++/5.3.0/tuple:856:5: note: candidate: template<class _Head, long unsigned int __i, class ... _Tail> constexpr _Head& std::__get_helper2(std::_Tuple_impl<__i, _Head, _Tail ...>&) 
    __get_helper2(_Tuple_impl<__i, _Head, _Tail...>& __t) noexcept 
    ^
/usr/local/include/c++/5.3.0/tuple:856:5: note: template argument deduction/substitution failed: 
/usr/local/include/c++/5.3.0/tuple:874:57: note: mismatched types 'std::integral_constant<long unsigned int, 0ul>()' and 'int' 
    { return std::forward<_Tp&&>(std::__get_helper2<_Tp>(__t)); } 
                 ^
/usr/local/include/c++/5.3.0/tuple:874:57: note: 'std::tuple<int>' is not derived from 'std::_Tuple_impl<__i, std::integral_constant<long unsigned int, 0ul>(), _Tail ...>' 
/usr/local/include/c++/5.3.0/tuple:861:5: note: candidate: template<class _Head, long unsigned int __i, class ... _Tail> constexpr const _Head& std::__get_helper2(const std::_Tuple_impl<__i, _Head, _Tail ...>&) 
    __get_helper2(const _Tuple_impl<__i, _Head, _Tail...>& __t) noexcept 
    ^
/usr/local/include/c++/5.3.0/tuple:861:5: note: template argument deduction/substitution failed: 
/usr/local/include/c++/5.3.0/tuple:874:57: note: mismatched types 'std::integral_constant<long unsigned int, 0ul>()' and 'int' 
    { return std::forward<_Tp&&>(std::__get_helper2<_Tp>(__t)); } 
                 ^
/usr/local/include/c++/5.3.0/tuple:874:57: note: 'std::tuple<int>' is not derived from 'const std::_Tuple_impl<__i, std::integral_constant<long unsigned int, 0ul>(), _Tail ...>' 

Answer 1

TL; DRは、私は両方のケースでgccと打ち鳴らすの行動が正しいと思います。

---

2つの呼び出しの間に微妙な違いがあります。私はただ説明するためにエイリアスを追加してみましょう：

using Zero = std::integral_constant<size_t, 0>; 
auto tuple = std::make_tuple(123); 

あなたが書いた：

std::get<Zero()>(tuple); 

Zero()がタイプです。 Zeroを返すnullary関数です。このように、std::getのバージョンはstd::get<T>()となります。 tupleにタイプがZero()の要素がないため、これはエラーです。あなたが書く一方

、：

Zero n; 
std::get<n>(tuple); 

nはタイプではありません - それは今までの値です。 std::integral_constantにはconstexpr operator size_t()が含まれているため、その番号が使用され、std::get<I>()と呼ばれることになります。

同じ単にもブレース初期化を使用することによって達成することができる：

std::get<Zero{}>(tuple); 

Zero{}は間違いタイプではないからです。