c++ - 我可以依靠空参数包来正确扩展吗？

Question

以下代码定义了函数 subst_first，它将一个整数数组的第一个元素替换为另一个数组的内容。它在 gcc 和 clang ( live demo ) 下工作。对于案例 //1、//2 和 //3 生成的 index_sequence's 之一是空的。因此，//4 处的参数包之一具有零个元素。这让我感到不安。我可以相信这种行为是符合标准的吗？

template<size_t n, size_t ... S1, size_t ... S2>
constexpr
std::array<int,3>
subst_first_impl( std::array<int,n> const &v1,
                  std::array<int,3> const &v2,
                  size_t min_n,
                  std::index_sequence<S1...>,
                  std::index_sequence<S2...> )
{    return std::array<int,3>{{ v1[S1]..., v2[min_n+S2]... }}; } // 4

template<size_t n>
constexpr
std::array<int,3>
subst_first( std::array<int,n> const &v1,
             std::array<int,3> const &v2 )
{   auto const min_n= std::min( size_t(3), n );
    return subst_first_impl( v1, v2, min_n,
                             std::make_index_sequence< min_n >(),
                             std::make_index_sequence< size_t(3) - min_n >() );
}

int main(){
    constexpr std::array<int,3>  a1{{1,2,3}};

    constexpr std::array<int,2>  b1{{4,5}};
    constexpr std::array<int,3>  b2{{6,7,8}};
    constexpr std::array<int,4>  b3{{9,10,11,12}};
    constexpr std::array<int,0>  b4{};

    constexpr auto b1a1= subst_first( b1, a1 );
    // ==> 4, 5, 3

    constexpr auto b2a1= subst_first( b2, a1 ); // 1
    // ==> 6, 7, 8

    constexpr auto b3a1= subst_first( b3, a1 ); // 2
    // ==> 9, 10, 11

    constexpr auto b4a1= subst_first( b4, a1 ); // 3
    // ==> 1, 2, 3
}

注意:我不是在寻找替换数组元素的解决方案。我对 index_sequence 和参数包的行为很感兴趣。

Answer 1

首先，std::make_index_sequence<0>完全有效(§20.5.3 [intseq.make]):

[ Note: make_integer_-sequence<int, 0> denotes the type integer_sequence<int> —end note ]

所以在你的例子中，你得到一个 std::index_sequence<size_t> .

并且根据 §14.5.3/7 [temp.variadic]，长度为 0 的包扩展的实例化是完全有效的:

The instantiation of a pack expansion that is neither a sizeof... expression nor a fold-expression produces a list E1, E2, ..., EN, where N is the number of elements in the pack expansion parameters. [...]

All of the E_i become elements in the enclosing list. [...] When N is zero, the instantiation of the expansion produces an empty list. Such an instantiation does not alter the syntactic interpretation of the enclosing construct, even in cases where omitting the list entirely would otherwise be ill-formed or would result in an ambiguity in the grammar.