按std :: variant中的类型获取索引

问题描述

标准库中是否有实用程序可在 std :: variant 类型的 index >？还是我应该自己做一个？也就是说，我想在 std :: variant< A，B，C> 中获得 B 的索引，然后有返回 1 。

Is there a utility in the standard library to get the index of a given type in std::variant? Or should I make one for myself? That is, I want to get the index of B in std::variant<A, B, C> and have that return 1.

有 std :: variant_alternative 用于相反的操作。当然，在 std :: variant 的列表上可能有许多相同的类型，因此此操作不是双向的，但对我来说不是问题（我可以在列表上首次出现类型，或者在 std :: variant 列表上具有唯一类型）。

There is std::variant_alternative for the opposite operation. Of course, there could be many same types on std::variant's list, so this operation is not a bijection, but it isn't a problem for me (I can have first occurrence of type on list, or unique types on std::variant list).

推荐答案

我找到了答案对于元组并对其进行了稍微修改：

I found this answer for tuple and slightly modificated it:

template<typename VariantType, typename T, std::size_t index = 0>
constexpr std::size_t variant_index() {
    if constexpr (index == std::variant_size_v<VariantType>) {
        return index;
    } else if constexpr (std::is_same_v<std::variant_alternative_t<index, VariantType>, T>) {
        return index;
    } else {
        return variant_index<VariantType, T, index + 1>();
    }
}

它对我有用，但是现在我很好奇如果作为结构，如何在没有constexpr的情况下以旧方式进行操作。

It works for me, but now I'm curious how to do it in old way without constexpr if, as a structure.

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