How to concatenate static strings at compile time?

Question

I am trying to use templates to create an analogue of the type_info::name() function which emits the const-qualified name. E.g. typeid(bool const).name() is "bool" but I want to see "bool const". So for generic types I define:

template<class T> struct type_name { static char const *const _; };

template<class T> char const *const type_name<T>::_ = "type unknown";

char const *const type_name<bool>::_ = "bool";
char const *const type_name<int>::_ = "int";
//etc.

Then type_name<bool>::_ is "bool". For non-const types obviously I could add a separate definition for each type, so char const *const type_name<bool const>::_ = "bool const"; etc. But I thought I would try a partial specialization and a concatenation macro to derive in one line the const-qualified name for any type which has its non-const-qualified name previously defined. So

#define CAT(A, B) A B

template<class T> char const *const type_name<T const>::_
    = CAT(type_name<T>::_, " const"); // line [1]

But then type_name<bool const>::_ gives me error C2143: syntax error: missing ';' before 'string' for line [1]. I think that type_name<bool>::_ is a static string known at compile time, so how do I get it concatenated with " const" at compile time?

I tried more simple example but same problem:

char str1[4] = "int";
char *str2 = MYCAT(str1, " const");

Answer 1

EDIT - See my new, improved answer here.

---

Building on @Hededes answer, if we allow recursive templates, then concatenation of many strings can be implemented as:

#include <string_view>
#include <utility>
#include <iostream>

namespace impl
{
/// Base declaration of our constexpr string_view concatenation helper
template <std::string_view const&, typename, std::string_view const&, typename>
struct concat;

/// Specialisation to yield indices for each char in both provided string_views,
/// allows us flatten them into a single char array
template <std::string_view const& S1,
          std::size_t... I1,
          std::string_view const& S2,
          std::size_t... I2>
struct concat<S1, std::index_sequence<I1...>, S2, std::index_sequence<I2...>>
{
  static constexpr const char value[]{S1[I1]..., S2[I2]..., 0};
};
} // namespace impl

/// Base definition for compile time joining of strings
template <std::string_view const&...> struct join;

/// When no strings are given, provide an empty literal
template <>
struct join<>
{
  static constexpr std::string_view value = "";
};

/// Base case for recursion where we reach a pair of strings, we concatenate
/// them to produce a new constexpr string
template <std::string_view const& S1, std::string_view const& S2>
struct join<S1, S2>
{
  static constexpr std::string_view value =
    impl::concat<S1,
                 std::make_index_sequence<S1.size()>,
                 S2,
                 std::make_index_sequence<S2.size()>>::value;
};

/// Main recursive definition for constexpr joining, pass the tail down to our
/// base case specialisation
template <std::string_view const& S, std::string_view const&... Rest>
struct join<S, Rest...>
{
  static constexpr std::string_view value =
    join<S, join<Rest...>::value>::value;
};

/// Join constexpr string_views to produce another constexpr string_view
template <std::string_view const&... Strs>
static constexpr auto join_v = join<Strs...>::value;


namespace str
{
static constexpr std::string_view a = "Hello ";
static constexpr std::string_view b = "world";
static constexpr std::string_view c = "!";
}

int main()
{
  constexpr auto joined = join_v<str::a, str::b, str::c>;
  std::cout << joined << '\n';
  return 0;
}

I used c++17 with std::string_view as the size method is handy, but this could easily be adapted to use const char[] literals as @Hedede did.

This answer is intended as a response to the title of the question, rather than the more niche problem described.