static auto constexpr t={“red”,“black”,“green”}是什么类型;派生到?

所以我假设类型是“initializer\u list”?

对。

有没有一种方法可以像:t[1]这样得到字符串“black”(假设索引从0开始)?

使用 std::array ,它与C++ 17类模板类型演绎很好地结合:

static auto constexpr t = std::array{"red", "black", "green"};

std::cout << t[1] << "\n"; // outputs 'black'

constexpr -如果将编译时常量传递给 operator[] 或 std::get .

constexpr const char* black = t[1];
constexpr const char* red = std::get<0>(t);

https://en.cppreference.com/w/cpp/utility/initializer_list

operator[] 成员函数,但是,唯一的访问器 std::initializer_list::begin std::initializer_list::end

#include <iostream>
#include <initializer_list>
#include <stdexcept>

namespace my {

template <typename T>
constexpr auto get(std::initializer_list<T> lst, std::size_t i)
{
    if ( i >= lst.size() )
        throw std::out_of_range("Out of bounds access.\n");
    return *(lst.begin() + i);   
}

template <std::size_t I, typename T>
constexpr auto get(std::initializer_list<T> lst) 
{
    // Note: https://stackoverflow.com/questions/5438671/static-assert-on-initializer-listsize
    if ( I >= lst.size() )
        throw std::out_of_range("Out of bounds access.\n");
    return *(lst.begin() + I);   
}

}

int main()
{
    // Note: https://stackoverflow.com/questions/16063123/is-it-legal-to-declare-a-constexpr-initializer-list-object
    static constexpr auto t = {"red", "black", "green"};

    std::cout << my::get(t, 1) << '\n' << my::get<2>(t) << '\n';
}