在lua中调用undefined函数时接收任何通知

class CWindowUI {
  public CWindowUI(const char* title,int width,int height);
  .....
};

static int CreateWindow(lua_State *l)
{
    int          width,height;
    char        *title;
    CWindowUI  **winp,*win;

    name = (char *) luaL_checkstring(l,1);
    width= lua_tounsigned(l,2);
    height= lua_tounsigned(l,3);

    win = new CWindowUI(title,width,height);
    if (win == NULL) {
        lua_pushboolean(l,0);
        return 1;
    }

    winp = (CWindowUI **) lua_newuserdata(l,sizeof(CWindowUI *));
    luaL_getmetatable(l,"WindowUI");
    lua_setmetatable(l,-2);
    *winp = win;

    return 1;
}

在Lua代码中：

local win = CreateWindow("title",480,320);
win:resize(800,600);

现在我的问题是：

函数CreateWindow将返回名为win的对象,并且未定义函数resize.当我在Lua中调用未定义的函数时如何收到通知？

通知应包括字符串“resize”和参数800,600.
我想修改源以将未定义的函数映射到回调函数,但它是不正确的.