Correct way to write C/Lua mixed module

[andre@...]

Hi All,

I'm writing a module which has mixed native C functions and Lua code. My goal is to have it available by only requiring it. My DLL exports only one function:

------------------------------8<------------------------------
extern "C"
{
  LUAMOD_API int luaopen_task( lua_State* L )
  {
    static const luaL_Reg statics[] =
    {
      { "getCurrentTimeMillis", module_getCurrentTimeMillis },
      { NULL, NULL }
    };

    // Register C functions.
    luaL_newlib( L, statics ); // table

    // Register Lua functions.
    if ( luaL_loadstring( L, s_LuaCode ) != LUA_OK ) // table function
    {
      fprintf( stderr, "%s\n", lua_tostring( L, -1 ) );
      return 0;
    }

    if ( lua_pcall( L, 0, 1, 0 ) != LUA_OK ) // table function
    {
      fprintf( stderr, "%s\n", lua_tostring( L, -1 ) );
      return 0;
    }

    lua_pushvalue( L, -2 ); // table function table@1

    if ( lua_pcall( L, 1, 0, 0 ) != LUA_OK ) // table
    {
      fprintf( stderr, "%s\n", lua_tostring( L, -1 ) );
      return 0;
    }

    return 1;
  }
}
------------------------------8<------------------------------

s_LuaCode is a C array generated with the following source code

------------------------------8<------------------------------
local function doSomething()
  -- ...
end

return function( module )
  module.doSomething = doSomething
end
------------------------------8<------------------------------

The Lua code returns a function which receives a table where the module functions are to be registered, and this function is called as part of luaopen_task.

Questions:

1. Is there a better way to handle a C/Lua mixed module? I don't need anything fancy, just register Lua and C functions.
2. Can/should I call lua_error in luaopen_task if one of the functions I call fails?

Thanks in advance,

Andre