Testcase for Luabridge Placing luabridge "registration" into different translation units can lead to types not being recognized correctly

A.cpp

#include "A.hpp"

#include <iostream>
#include <string>

#include <LuaBridge/LuaBridge.h>

std::string A::getName() const { return std::string("A"); }

void A::register_lua(lua_State *L)
{
    using namespace luabridge; 
    getGlobalNamespace (L)
        .beginNamespace ("test")
            .beginClass <A> ("A")
                .addConstructor<void(*)(void)>()
                .addProperty("name", &A::getName)
            .endClass ()
        .endNamespace ();
}

A.hpp

#ifndef A_H
#define A_H

#include <string>

#include <lua5.2/lua.hpp>

struct A {
    std::string getName() const;
    static void register_lua(lua_State *L);
};

#endif

main.cpp

#include <iostream>
#include <string>

#include <lua5.2/lua.hpp>
#include <lua5.2/lualib.h>

#include <LuaBridge/LuaBridge.h>

#include "A.hpp"

void printA(A *a)
{
    std::cout << a->getName() << std::endl;
}

int main(int argc, char *argv[]) {
    int iarg;
    lua_State *L = luaL_newstate();

  luaL_openlibs(L);
	
    A::register_lua(L);

	{
		using namespace luabridge; 
		getGlobalNamespace (L)
			.addFunction ("printA", &printA);
	}

    int s = luaL_dostring(L,
        "a = test.A()\n"
        "printA(a)\n"
    );
    if(s != 0) {
        std::cerr << "Error: " <<  lua_tostring(L, -1) << std::endl;
        lua_pop(L, 1);
    }

    lua_close(L);

    return 0;
}