c++ - 如何将类链接到枚举？

Question

我有以下代码:

#include <string>

enum class Hobbit {
    // typedef HobbitHelper helper;
    UNKNOWN = -1, Bilbo, Frodo, Samwise
};

struct HobbitHelper {
    static Hobbit decode(std::string const& s) {
        if (s == "Bilbo") {
            return Hobbit::Bilbo;
        }
        else if (s == "Frodo") {
            return Hobbit::Frodo;
        }
        else if (s == "Samwise") {
            return Hobbit::Samwise;
        }
        else {
            return Hobbit::UNKNOWN;
        }
    }
};

enum class Wizard {
    // typedef Apprentice helper;
    UNKNOWN = -1, Gandalf, Radagast, Saruman
};

struct Apprentice { // WizardHelper :)
    static Wizard decode(std::string const& s) {
        if (s == "Gandalf") {
            return Wizard::Gandalf;
        }
        else if (s == "Radagast") {
            return Wizard::Radagast;
        }
        else if (s == "Saruman") {
            return Wizard::Saruman;
        }
        else {
            return Wizard::UNKNOWN;
        }
    }
};

template <typename T>
T
decoder(std::string s)
{
    return ??::decode(s);
    // if the typedefs were allowed, I could use T::helper::decode()
}

int main()
{
    std::string s{ "Rincewind" };

    auto h = decoder<Hobbit>(s);
    auto w = decoder<Wizard>(s);
}

如何安排在decoder 中调用适当的助手类(HobbitHelper 或 Apprentice)？我不能在枚举中声明嵌套类型，就好像它是一个类一样。我还尝试从枚举派生助手(因为助手本身没有数据)，但这也是不允许的。

有什么想法吗？

Answer 1

你可以让 helper 类型特征是外部的，并在枚举类型上模板化，为每个 enum 明确特化:

template <typename T> struct type_is { using type = T; };

template <typename > struct helper;

template <> struct helper<Hobbit> : type_is<HobbitHelper> { };
template <> struct helper<Wizard> : type_is<Apprentice> { };

template <typename T>
using helper_t = typename helper<T>::type;

然后 decode 将访问它:

template <typename T>
T decoder(std::string s)
{
    return helper_t<T>::decode(s);
}