更新：已编辑以修复工作示例中的编译问题。
我想要做类似以下的事情，以便该函数可以接受枚举类实例的列表或保存它们的结构体，但是“auto myFunc2<EList<Types...>>”的定义失败，出现“期望常数而不是E::A”的错误。
```cpp
#include <cstdint>
#include <iostream>
enum class E { A = 0, B = 1 };
template <E... Types> struct tl2 {};
using my_list_2 = tl2<E::A>;
template <E... Types>
auto myFunc2 = [] {
    std::cout << "Size: " << sizeof...(Types) << std::endl;
};
template <template <E...> typename EList, E... Types>
auto myFunc2<EList<Types...>> = [] {
    std::cout << "Size: " << sizeof...(Types) << std::endl;
};
int main() {
    myFunc2<E::A, E::B>();  // 此调用打印类型列表的大小
                          // 当注释掉myFunc2<Elist<Types..>>时可以正常工作
    // myFunc2<my_list_2>();      // 这会导致错误
}

#include <cstdint>
#include <iostream>
template <typename ... Types> struct tl
{
};
using my_list = tl <int, float, uint64_t>;
template <typename ... Types>
static constexpr auto myFunc2 = [] {
    std::cout << "Size: " << sizeof...(Types) << std::endl;
};
template <template <class...> class TL, typename ... Types>
static constexpr auto myFunc2<TL<Types...>> = [] {
    std::cout << "Size: " << sizeof...(Types) << std::endl;
};
int main() {
    myFunc2<int, uint64_t, bool, uint16_t>();
    myFunc2<my_list>();
}

真正的函数模板可以做到这一点（通常需要额外的辅助来推导参数包）：
```cpp
template <typename... Types>
void myFunc2() {
    std::cout << "Size: " << sizeof...(Types) << std::endl;     
}
namespace detail {
template <template<typename...> typename EList, typename... Types>
void myFunc2(EList<Types...>) {
    std::cout << "Size: " << sizeof...(Types) << std::endl;     
}
}
template <typename EList>
void myFunc2() {
    detail::myFunc2(EList());
}

<details>
<summary>英文:</summary>
True **function** templates can do this (with the usual extra helper for deducing a pack):

This approach allows “one template” to accept template arguments of different **kinds** because the templates with the wrong kinds of template parameters will simply be ignored during overload resolution.
</details>