How to pretty print the name of a template parameter at compile time - c++

Question

The question is rather simple: how to pretty print the name of a template parameter in a C++ class and assign it to a class variable at compile time ?
It seems that both typeinfo (typeid) and boost::typeindex must be evaluated at runtime or as least some part of them. This apparently does not allow the compiler to completely solve a constexpr containing a call to one of this function.
template<typename T>
class X
{
public:
static const char * name = /* SOME C++ code transforming T in a string (either std::string or char */
};
What am I missing ?
Is it only possible to generate a name at runtime ? In that case, does I really need an instantiated object ? It doesn't seem right to me, because the following perfectly work without any instance:
#include <iostream>
#include <string>
#include <boost/type_index.hpp>
using namespace std;
template<class T>
class X
{
public:
static std::string name()
{
return boost::typeindex::type_id<T>().pretty_name();
}
};
struct foobar {};
int main()
{
cout << X<int>::name() << endl;
cout << X<foobar>::name()<< endl;
}
So instead of having name() as a class method, I'd like to have it as a class variable.

Answer 1

I think, it is possible to use custom Type Traits. Please see the next example:
#include <iostream>
#include <string>
using namespace std;
//Using stub type traits
template <class T>
struct TypeTraits;
//your TypeTraits for specific types...
template<>
struct TypeTraits<int>
{
constexpr static const char *name = "int";
};
template<class T>
class X
{
public:
constexpr static const char * name = TypeTraits<T>::name;
};
struct foobar {};
//TypeTraits for custom foobar
template<>
struct TypeTraits<foobar>
{
constexpr static const char *name = "foobar";
};
int main()
{
//Now you can use static member here
cout << X<int>::name << endl;
cout << X<foobar>::name<< endl;
}
Also TypeTraits can be used (and expanded) for other purposes.