是否可以將完整的“路徑”保存到C函數中的Lua表字段？

Question

如果我在Lua中有這樣的全局對象：

global_object = { }
global_object.stat_group_1 = { }
global_object.stat_group_2 = { }
global_object.stat_group_3 = { }
global_object.stat_group_1.stat_1 = 1 -- value changes with time
global_object.stat_group_1.stat_2 = 2 -- value changes with time
global_object.stat_group_1.stat_3 = 3 -- value changes with time
-- ... and same thing for other stat_groups

我的問題是關於luaL_ref ， lua_rawgeti和lua_getfield 。 我可以使用luaL_ref來保存每個stat的路徑，以避免在堆棧上顯式調用所有stat ，如下所示：

int global_object_ref;
int stat_group_1_ref;
int stat_1_ref;

//assume this function has been called before any of the get_* functions
int start ( lua_State * L )
{
    lua_getfield ( L, LUA_RIDX_GLOBALS, "global_object" );
    lua_pushvalue ( L, -1 );
    global_object_ref = LuaL_ref ( L, LUA_REGISTRYINDEX );

    lua_getfield ( L, -1, "stat_group_1" );
    lua_pushvalue ( L, -1 );
    stat_group_1_ref = LuaL_ref ( L, LUA_REGISTRYINDEX );

    lua_getfield ( L, -1, "stat_1" );
    lua_pushvalue ( L, -1 );
    stat_group_1_ref = LuaL_ref ( L, LUA_REGISTRYINDEX );

    return 0;
}

//this is the prefered option. I would like this to be possible
int get_stat1_v1 ( lua_State * L )
{
    //stat_1 can have different values in the Lua table at different moments
    lua_rawgeti ( L, LUA_REGISTRYINDEX, stat_1_ref );

    //is this the value of the field stat_1?
    int value_of_stat_1 = lua_tointeger ( L, -1 );

    return 1;
}

//this is an alternative, in case v1 doesn't work. Would this work?
//again, remember stat_1 can have different values at different moments.
int get_stat1_v2 ( lua_State * L )
{
    lua_rawgeti ( L, LUA_REGISTRYINDEX, global_object_ref );
    lua_rawgeti ( L, LUA_REGISTRYINDEX, stat_group_1_ref );
    lua_rawgeti ( L, LUA_REGISTRYINDEX, stat_1_ref );

    //is this the value of the field stat_1?
    int value_of_stat_1 = lua_tointeger ( L, -1 );

    return 1;
}

請注意， v2將所有已保存的引用調用到堆棧上。 那樣有用嗎？

編輯：根據@Nicol Bolas的回答，我想提出一個v3 。 如果表和子表從不進行垃圾回收，但是它們的值不斷更新（想象一下子表的整個結構像一棵樹，每個子表是一個分支，每個基本值是一個葉子。樹的結構保持不變在執行期間相同，但葉子得到更新）。

//this is the v3, where I learned how the Lua registry interacts with C, and propose
//a direct leaf access, but indirect branch access.
int get_stat1_v3 ( lua_State * L )
{
    //can I skip the line bellow and go directly to the next? Or do I have
    //to follow the whole hierarchy of branches?
    lua_rawgeti ( L, LUA_REGISTRYINDEX, global_object_ref );//is this needed?

    //I know I'm repeating myself, but I want to know if the call above
    //is necessary or not. Can I directly cut to the chase by calling this function?
    lua_rawgeti ( L, LUA_REGISTRYINDEX, stat_group_1_ref );

    //Notice that for this to work I'd have to removed the reference to stat_1
    //from the registry of my proven flawed implementation, so I wouldn't
    //freeze and pin down the value of stat_1, which means stat_1_ref gets
    //removed from the code.
    lua_getfield ( L, -1, "stat_1" );

    //is this the dynamically up to date value of the field stat_1?
    int value_of_stat_1 = lua_tointeger ( L, -1 );

    return 1;
}

Answer 1

您可以這樣做，但是基本上您將放棄任何形式的垃圾收集。 注冊表是Lua狀態的一部分，因此只要這些表在注冊表中，它們就必須存在。 因此，它們所引用的任何對象都將存在，直到您注銷它們或關閉Lua狀態為止。

您並沒有真正保存“路徑”。 您將在這些位置的表中保存實際值。 因此，如果表的值發生更改，則不會更新注冊表中存儲的內容。 保存stat_1的值將保存為當前值，而不是可能更改為的值。