Question

我在这里拉我的头发，我可以在Lua和c ++之间共享一个数组，我甚至可以在Lua中创建一个对象数组（使用下面的代码）并访问它们的成员函数（例如obj [10]：setPosition （0,0,0））。我不能做的是将c ++对象发送到已经存在的Lua并让Lua调用它们各自的成员函数。例如：

objects = Scene.getAllObjects()
objects[5]:setPosition(0,0,0)

...不起作用，但是，下面的代码可以使用

for i=1,10 do
  objects[i] = Object.new("Box")
  objects[i]:setPosition(0,0,0)
end

...调用下面的c ++函数

int luaAddNewEditableObject(lua_State * L)
{
    const char * name = luaL_checkstring(L, 1);

    EditableObject ** udata = (EditableObject **)lua_newuserdata(L, sizeof(void*));
    EditableObject *obj = scene->addNewEditableObject(name);
    *udata = obj;

    luaL_getmetatable(L, "luaL_EditableObject");

    lua_setmetatable(L, -2);

    return 1;
}

所以基本上，如果它是在Lua中创建的那么没有问题，但如果c ++中已经存在对象，那么我需要将它放入Lua表/数组中，以便Lua可以为它们做魔术

请帮助

struct ObjectArray 
{
    int size;
    EditableObject *objects;  /* The cpp objects */
};

static int getAllObjects (lua_State *L) 
{
    //This creates the metatable for array notation
    size_t nbytes = sizeof(ObjectArray) + numObjects * sizeof(ObjectArray*);
    ObjectArray *objectArray = (ObjectArray*)lua_newuserdata(L, nbytes);
    objectArray->size = numObjects;

    for (int i = 0; i < numFoos; i++)
    {
            //This sets the c++ pointers to the lua_newuserdata 
                objectArray->objects[i] = objects[i];

        //So maybe here I need to assign the 'luaL_EditableObject' metatable for each object
        //so I can call it's member functions ??
    }
    luaL_getmetatable(L, "ObjectArray");
    lua_setmetatable(L, -2);

    return 1; 
}

Answer 1

好的，我想出来了，如果有其他人遇到这个问题，我会回答我自己的问题

extern "C"
{
#include <lua.h>
#include <lauxlib.h>
#include <lualib.h>
}

#include <map>

class Object 
{
public:
    Object()
    {
        x = 0;
        y = 0;
    }
    Object(int x, int y)
    {
        this->x = x;
        this->y = y;
    }
    int x, y;
};


std::map<std::string,Object> objects;

static int luaGetObjects(lua_State *L)
{
    lua_newtable(L);
    int i=0;
    for (auto it = objects.begin(); it != objects.end(); it++, i++) 
    {

        lua_pushstring(L, it->first.c_str());

        luaL_getmetatable(L, "luaL_Object");
        lua_setmetatable(L, -2);

        lua_rawseti(L, -2, i+1);
        stackDump(L);
    }

    return 1;   
}

static int luaSetPosition(lua_State* L) 
{
    const char* key = luaL_checkstring(L,1);

    int x = luaL_checknumber(L,2);
    int y = luaL_checknumber(L,3);

    objects[key].x = x;
    objects[key].y = y;

    return 0; 
}

static int luaGetPosition(lua_State* L) 
{
    const char* key = luaL_checkstring(L,1);

    lua_pushnumber(L, objects[key].x);
    lua_pushnumber(L, objects[key].y);

    return 2; 
}

void registerObject(lua_State *L)
{
    luaL_Reg regs[] =
    {
        { "setPosition", luaSetPosition },
        { "getPosition", luaGetPosition },
        { NULL, NULL }
    };

    luaL_newmetatable(L, "luaL_Object");
    luaL_register(L, NULL, regs);
    lua_pushvalue(L, -1);
    lua_setfield(L, -1, "__index");
}

int main()
{
    lua_State * L = luaL_newstate();
    luaL_openlibs(L);

    lua_pushcfunction(L, luaGetObjects);
    lua_setglobal(L, "getObjects");

    registerObject(L);

    objects["id001"] = Object(1,2);
    objects["id002"] = Object(3,4);
    objects["id003"] = Object(5,6);

    int erred = luaL_dofile(L, "hello.lua");
    if(erred)
        std::cout << "Lua error: " << luaL_checkstring(L, -1) << std::endl;

    lua_close(L); 

    return 0;
}

'hello.lua'代码：

objects = getObjects()

for i=1,#objects do
   x,y = objects[i]:getPosition();
   print(string.format("object[%d] x = %d y = %d",i,x,y))

   objects[i]:setPosition(x*100, y*100);

   x,y = objects[i]:getPosition();
   print(string.format("object[%d] x = %d y = %d after mul",i,x,y))
end

Lua可以访问并调用成员函数的c ++对象数组

1 个答案: