Question

GCC（5.3）＆amp; Clang（3.8）声称test中的第一行很糟糕，但第二行很好。 MSVC（2015.2）说，两者都是无效的。

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
}

根据标准，究竟是什么区别？这段代码是否合法？

第一行的clang错误：

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错误：

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 我认为gcc和clang在这两种情况下的行为都是正确的。

两次通话之间存在细微差别。我只想添加一个别名来说明：

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

当你写：

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

Zero()是类型。这是一个返回Zero的无效函数。因此，您调用的std::get版本采用的类型为std::get<T>()。由于tuple中没有Zero()类型的元素，因此这是一个错误。

另一方面，当你写：

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

n不是一种类型 - 它只是一种价值。由于std::integral_constant有constexpr operator size_t()，因此会使用std::get<I>()，并最终调用std::get<Zero{}>(tuple);，这会达到您所期望的效果。

同样可以通过简单地使用大括号初始化来实现：

Zero{}

因为{{1}}绝对不是一种类型。

constexpr模板参数古怪

1 个答案: