是否可以将变体的索引作为 constexpr 变量获取？

Question

我有以下代码 (play with example). It checks what the underlying value of a variant is and receives this underlying value using get。如您所见，代码变得非常重复。

#include <variant>
#include <string>
#include <iostream>
#include <vector>

int main()
{
    std::variant<int, double, std::string> v;
    // v holds a string
    v = "hi there!";

    switch(v.index()) {
        case 0: 
            {   // brackets to avoid cross initialization
                int i = std::get<int>(v);
                // handle int
            }    
            break;
        case 1: 
            {
                double d = std::get<double>(v);
                // handle double
            }
            break;
        case 2: 
            {
                std::string s = std::get<std::string>(v);
                // handle string
            }
            break;
    }
}

问题： 有什么方法可以将变体的索引作为 constexpr 变量吗？我想做这样的事情：

// what I would like to do
// auto val_of_v = std::get<v.index()>(v);

// Error: template argument deduction/substitution failed
//        in 'constexpr' expansion of v<...>::index()

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Answer 1

仅当变体是 constexpr 本身时：

constexpr std::variant<int, double> v{24};
auto a = std::get<v.index()>(v);

否则信息只能在运行时才知道

Answer 2

您正在寻找的是 std::visit，它是为您实现该转换的图书馆设施：

std::variant<int, double, std::string> v;
// v holds a string
v = "hi there!";

std::visit([](auto&& x){
    // x is whatever the variant was holding (could be int, double, or string)
}, v);

是否可以将变体的索引作为 constexpr 变量获取？

Is it possible to get the index of a variant as a constexpr variable?

c++

switch-statement

c++17

constexpr

variant