模板参数推导和默认模板参数

Question

下面的代码编译并按照我想要的方式使用clang和gcc,但在Visual Studio 2015 RTM中出错.我认为clang和gcc是正确的,但我不确定.这段代码应该编译吗？

#include <iostream>
#include <type_traits>
#include <utility>

template <typename T, size_t N, typename IS = decltype(std::make_index_sequence<N>{})>
struct Vector {
    T e_[N];
};

template <typename T, typename U, size_t N, size_t... Is>
constexpr auto operator+(const Vector<T, N, std::index_sequence<Is...>>& x,
                         const Vector<U, N>& y) {
    using V = std::common_type_t<T, U>;
    return Vector<V, N>{x.e_[Is] + y.e_[Is]...};
}

int main() {
    const auto v0 = Vector<float, 4>{1, 2, 3, 4};
    const auto v1 = Vector<float, 4>{5, 6, 7, 8};
    const auto v2 = v0 + v1;
    for (auto x : v2.e_) std::cout << x << ", ";
    std::cout << std::endl;
}

如果我将operator +更改为:Visual Studio将编译为ok:

template <typename T, typename U, size_t N, size_t... Is>
constexpr auto operator+(const Vector<T, N, std::index_sequence<Is...>>& x,
                         const Vector<U, N, std::index_sequence<Is...>>& y);

但我认为没有必要std::index_sequence<Is...>再次提出y.

Answer 1

IS不应该是这种类型的一部分Vector.相反,使用辅助函数:

template <typename T, std::size_t N>
struct Vector {
    T e_[N];
};

template <typename T, typename U, std::size_t N, std::size_t... Is>
inline constexpr auto add_impl(const Vector<T, N>& x, const Vector<U, N>& y,
                               std::index_sequence<Is...>) {
    using V = std::common_type_t<T, U>;
    return Vector<V, N>{x.e_[Is] + y.e_[Is]...};
}

template <typename T, typename U, std::size_t N,
          typename Is = std::make_index_sequence<N>>
constexpr auto operator+(const Vector<T, N>& x, const Vector<U, N>& y) {
    return add_impl(x, y, Is());
}

我无法访问VS2015来测试它,但它应该可以工作.