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

Question

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* c         


        

Answer 1

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

#using 
// 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
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
constexpr std::string_view type_name() {

    constexpr std::string_view funcsig =  detail::templated_function_name_getter();

    // 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.