获取类型列表大小的泛型实现

如何实现泛型 SizeOfT 排版模板?我正在学习C++模板元编程,决定实现 尺寸英尺

template <typename... Ts>
struct TypeList1;

template <typename... Ts>
struct TypeList2;

template <typename H, typename... Ts>
struct SizeOfT;

// Specialized for TypeList1
template <typename H, typename... Ts>
struct SizeOfT <TypeList1<H, Ts...>> {
    constexpr static auto value = 1 + sizeof...(Ts);
};

template <>
struct SizeOfT <TypeList1<>> {
    constexpr static auto value = 0;
};

// Specialized for TypeList2, works fine but
// it would be less code if generic SizeOfT can be implemented which can handle
// both TypeList1 and TypeList2 and maybe any future TypeList3 and so on...
template <typename H, typename... Ts>
struct SizeOfT <TypeList2<H, Ts...>> {
    constexpr static auto value = 1 + sizeof...(Ts);
};

template <>
struct SizeOfT <TypeList2<>> {
    constexpr static auto value = 0;
};

int main() {
    using tl1 = TypeList1<int, char, bool>;
    using tl2 = TypeList2<float, double>;

    static_assert(SizeOfT<tl1>::value == 3, "tl1 size is not 3");
    static_assert(SizeOfT<tl2>::value == 2, "tl2 size is not 2");

    return 0;
}

在上面的代码中,一切正常。但是,我想 更通用,这样如果新的类型列表 TypeList3 是的,我不需要为此提供任何专门化。

我使用的是C++ 11兼容编译器。