Lua C API代码不调用__newindex函数但调用其他函数

时间:2012-11-28 18:47:19

标签: c indexing lua metatable

我编写了一些代码来将自定义函数和__newindex__index函数分别注册到2个独立的函数中。我的代码的目标是让Lua脚本编写器可以看到基于特异性子级组织的函数和变量。例如,用户可以使用以下命令:

orc.chief.attack();
orc.chief.flee();
orc.chief.hp = 100;
orc.pawn.attack();
elf.wood.attack();
elf.wood.hp = 200;

所以基本上是一个有2层,然后是函数调用或变量的系统。如果我正确理解Lua,那就是表格中的表格。当用户设置变量时,它应该跳转__newindex调用(不仅要处理设置值,还要访问将通过电机设置动画的物理对象)。我还假设表orc中的主表只是看到分配给它的许多函数,无论它是attack还是__newindex。为了便于在代码开发时添加新的变量和函数,我创建了两个函数:一个用于创建一个函数,另一个用于创建一个变量。函数create只是注册函数,变量create只是创建一个新的表元素并注册__newindex__index的函数。以下是代码:

int orcChiefhp;

luaL_Reg Orc_Module[] = {
    {"attack", OrcAttack},
    {"flee", OrcFlee},
    {NULL, NULL}};

const luaL_Reg orcChief_metareg[] = {
    {"__index", orcChief__index},
    {"__newindex", orcChief__newindex},
    {NULL, NULL}};

int OrcAttack(lua_State *L)
{
  //code to cause the motors to swing the weapon...
  return 0;//0 parameters come back as the data
}

int orcChief__newindex(lua_State *L)
{
const char *idx;
    if(lua_isstring(L,2))
    {
        idx = lua_tostring(L,2);//gets the string so we can get the variable of the struct
        if(strcmp(idx, "hp")==0)
        {
            lua_pushnumber(L, orcChiefhp);
        }
        else
            lua_pushnil(L);
    }
    return 1;
}

void registerFunctions(lua_State *L, const char *libname, const char *sublibname, const luaL_Reg *funcs)
{
int isitnil;

    lua_getglobal(L, libname);
    isitnil = lua_isnil(L, -1);
    if(isitnil)
    {
        lua_pop(L, 1);
        lua_newtable(L);    // create 'libname' table
    }
    // no sublib: just import our library functions directly into lib and we're done
    if (sublibname == NULL)
    {
         luaL_setfuncs(L, funcs, 0);
    }
    // sublib: create a table for it, import functions to it, add to parent lib
    else
    {
         lua_newtable(L);
         luaL_setfuncs(L, funcs, 0);
         lua_setfield(L, -2, sublibname);
    }
    if(isitnil)
         lua_setglobal(L, libname);//this will pop off the global table.
    else
         lua_pop(L, 1);//the global table is still on the stack, pop it off
}

void registerIntegerVariable(lua_State *L, const char *libname, const char *sublibname, const char *variableName,
    const char *metatableName, const luaL_Reg *metatableFuncs, int defaultValue)
{
int isLibnameNil;
int isSubnameNil;
    lua_getglobal(L, libname);//get the libname
    isLibnameNil = lua_isnil(L, -1);//check to see if it exists
    if(isLibnameNil)//if it doesn't exist, create a new one
    {
        lua_pop(L, 1);//pop off the nil
        lua_newtable(L);    // create 'libname' table
    }

    // no sublib: just import our library functions directly into lib and we're done
    if (sublibname == NULL)//if we want the functions at the lib level then just set the functions
    {
        lua_pushstring(L, variableName);//push the variable name
        lua_pushnumber(L, defaultValue);//push the default value on the stack
        lua_rawset(L, -3);//add the variable to the table (rawset is like settable but doesn't call __index)
        luaL_newmetatable(L, metatableName);//create the metatable
        luaL_setfuncs(L, metatableFuncs, 0);//set the metatable functions for __newindex and __index
        lua_setmetatable(L, -2);//set the metatable to the libtable
    }
    // otherwise we need to create a table for the sublibname, import functions to it, add to parent lib.
    else
    {
        lua_getfield(L, -1, sublibname);//see if the sublibname is under the global libname
        isSubnameNil = lua_isnil(L, -1);//is it a nil
        if(isSubnameNil)//if it is, then we need to create the sublibname
        {
            lua_pop(L, 1);//pop off the nil
            lua_newtable(L);//creates the new sublibname table 
        }
        lua_pushstring(L, variableName);//push the variable name
        lua_pushnumber(L, defaultValue);//push the default value on the stack
        lua_rawset(L, -3);//add the variable to the table and push it (rawset is like settable but doesn't call __index)
        luaL_newmetatable(L, metatableName);//create the metatable
        luaL_setfuncs(L, metatableFuncs, 0);//add the metamethods
        lua_setmetatable(L, -2);//set the metatable to the sublibname
        if(isSubnameNil)
          lua_setfield(L, -2, sublibname);//now we need to add the sublibname to the libname
    }

    if(isLibnameNil)
        lua_setglobal(L, libname);//set the global name if it was new
    else
        lua_pop(L, 1);
}

然后,在我的main()中,我调用了这样的函数:

execContext = luaL_newstate();
//adding lua basic library
luaL_openlibs(execContext);

//now register all the functions with Lua
registerFunctions(execContext, "orc", "chief", Orc_Module);
registerFunctions(execContext, "orc", "pawn", Orc_Module);
registerFunctions(execContext, "elf", "wood", Elf_Module);
//now register all the variables with Lua
registerIntegerVariable(execContext, "orc", "chief", "hp", "chief_meta", orcChief_metareg, 0);

当我在Lua脚本中运行代码和泵时,orc.chief.attack()调用我的OrcAttack()函数,但orc.chief.hp = 100从不调用我的orcChief__newindex()函数。我甚至已经注释了registerFunctions次来电,以防它们以某种方式干扰,只有registerIntegerVariable本身仍然不会触发orcChief__newindex().任何想法?

1 个答案:

答案 0 :(得分:2)

在表中设置字段时,不会调用

__newindex。在表中设置 new 字段时会调用它。如果该字段已存在,则不会调用__newindex

如果希望为表上的每个设置操作调用__newindex,则不能允许设置操作实际修改该表。这通常通过创建用户使用的称为代理表的空表来完成。代理表实际上是空的,必须始终保持不变;拦截所有get和set调用,将它们传递给用户从未看到的无法访问的内部表。

或者您使用一些userdata而不是表。总是要求__newindex