Can I obtain C++ type names in a constexpr way?

问题

I would like to use the name of a type at compile time. For example, suppose I've written:

constexpr size_t my_strlen(const char* s)
{
        const char* cp = s;
        while(*cp != '\0') { cp++; };
        return cp - s;
}

and now I want to have:

template <typename T>
constexpr auto type_name_length = my_strlen(typeid(T).name());

But alas, typeid(T).name() is just const char*, not constexpr... is there some other, constexpr way to get a type's name?

回答1:

Well, you could, sort of, but probably not quite portable:

struct string_view
{
    char const* data;
    std::size_t size;
};

inline std::ostream& operator<<(std::ostream& o, string_view const& s)
{
    return o.write(s.data, s.size);
}

template<class T>
constexpr string_view get_name()
{
    char const* p = __PRETTY_FUNCTION__;
    while (*p++ != '=');
    for (; *p == ' '; ++p);
    char const* p2 = p;
    int count = 1;
    for (;;++p2)
    {
        switch (*p2)
        {
        case '[':
            ++count;
            break;
        case ']':
            --count;
            if (!count)
                return {p, std::size_t(p2 - p)};
        }
    }
    return {};
}

And you can define your desired type_name_length as:

template <typename T>
constexpr auto type_name_length = get_name<T>().size;

DEMO (works for clang & g++)

回答2:

(Based on @melak47's gist and using C++17):

#using <string_view>
// and if it's not C++17, take the GSL implementation or just roll your own struct - 
// you only need to implement 3 or 4 methods here.

namespace detail {

template<typename T>
constexpr const char* templated_function_name_getter() {
#if defined(__GNUC__) || defined(__clang__)
    return __PRETTY_FUNCTION__;
#elif defined(_MSC_VER)
    return __FUNCSIG__;
#else
#error unsupported compiler (only GCC, clang and MSVC are supported)
#endif
}

} // namespace detail

template<typename T>
constexpr std::string_view type_name() {

    constexpr std::string_view funcsig =  detail::templated_function_name_getter<T>();

    // Note: The "magic numbers" below 

    #if defined(__GNUC__) || defined(__clang__)
    constexpr auto start_bit = std::string_view{"T = "};
    constexpr auto end_bit = std::string_view{"]"};
    #elif defined(_MSC_VER)
    constexpr auto start_bit = std::string_view{"detail::templated_function_name_getter<"};
    constexpr auto end_bit = std::string_view{">("};
    #else
    #error unsupported compiler (only GCC, clang and MSVC are supported)
    #endif

    constexpr auto start = funcsig.find(start_bit);
    constexpr auto end = funcsig.rfind(end_bit);
    static_assert(
        start != funcsig.npos and end != funcsig.npos and end > start, 
        "Failed parsing the __PRETTY_FUNCTION__/__FUNCSIG__ string");
    }
    return funcsig.substr(start + start_bit.size(), end - start - start_bit.size());
}

Note: There's now a nicer version of this approach implemented here.

来源：https://stackoverflow.com/questions/35941045/can-i-obtain-c-type-names-in-a-constexpr-way