而不是在 lua 中创建一个对象，如何让 lua 直接高C++对象来启动方法?

经过大量的研究，我终于找到了一个解决方案，尽管我不确定它的效率如何，但它实际上正在按照我想要的方式工作。
所以最终的解决方案是，注册一个从我需要的指针转换为lua的intptr_t，(有关intptr_t的更多信息，请单击此处(主要出于安全原因，您可以将其视为一种仅用于指针的int设计))并在我使用它时将其传递给静态函数

以下是修改后的代码：

在Test_Class.cpp，之后：

luabind::module(myLuaState)[
.def("shoot_a_bullet", &shoot_a_bullet)
];// since the functions are static now it will not need the namespace

我已经注册了指针的intptr_t版本：

//first convert pointer to intptr_t
intptr_t stage = (intptr_t)stg;//stg is the pointer
//then register it to lua
//if you try to register a pointer it will give you runtime error
luabind::globals(myLuaState)["stage"] = stage;

我们还需要稍微改变一下函数：首先，我们需要让函数静态，所以在头文件中：

//change the line defined the function to
void static shoot_a_bullet(intptr_t stg, int damage);

现在我们必须用它自己的函数做一些事情：

void Test_Class::shoot_a_bullet(intptr_t stg, int damage)
{
//we need to convert it back to pointer first, so
Test_Stage* stage = (Test_Stage*)stg;//this is the part I'm not sure about efficiency 
stage->create_bullet(damage, "wowo");
}

所以现在使用 lua 的代码将非常简单： shoot_a_bullet(舞台，134);

我们完成了！生活是美好的！这些东西花了我 3 周的时间。

你可能不明白我说的话，这是我写的完整测试代码：

main.cpp：

#include <iostream>
#include "test_stage.h"
using namespace std;
int x;
int main() {
cin >> x;
cin.get();// clean the n(when you press enter) after the number
Test_Stage stage = Test_Stage(x);
}

Test_Stage.h：

class Test_Class;
#ifndef TEST_STAGE_H
#define TEST_STAGE_H
extern "C"
{
#include <lua.h>
#include <lualib.h>
#include <lauxlib.h>
}
#include ".luabindluabind.hpp"
#include <iostream>
#include "test_class.h"
using namespace std;
class Test_Stage
{
public:
int x;
Test_Stage();
Test_Stage(int num);
void create_bullet(int damage, string name); /*This is currently useless
before i have understand how to include each other using foward referance*/
void create_class(int num);
Test_Class t_class;
~Test_Stage();
private:
};
#endif

Test_StageCPP：

#include"test_stage.h"
#include "test_class.h"// and as you see i included both files
Test_Stage::Test_Stage()
{
}
Test_Stage::Test_Stage(int num)
{
x = num;// a variable specific in this object(to test pointer)
create_class(num);
}
void Test_Stage::create_bullet(int damage, string name)
{
cout << "created damage: " << damage << " to " << x << endl;/*using the value
created in this object to see if pointer is actually working*/
}
void Test_Stage::create_class(int num)
{
Test_Class t_class = Test_Class(this, num);
}
Test_Stage::~Test_Stage()
{
}

Test_Class.h：

#ifndef TEST_CLASS_H    
#define TEST_CLASS_H
extern "C"
{
#include <lua.h>
#include <lualib.h>
#include <lauxlib.h>
}
#include ".luabindluabind.hpp"
#include <iostream>
using namespace std;
class Test_Stage;
class Test_Class
{
public:
int x;
Test_Class();
Test_Class(Test_Stage* stage, int num);
void static shoot_a_bullet(intptr_t stg, int damage);
~Test_Class();
private:
};
#endif TEST_CLASS_H

最后，Test_Class.cpp：

#include "test_class.h"
#include"test_stage.h"
void Test_Class::shoot_a_bullet(intptr_t stg, int damage)
{
Test_Stage* stage = (Test_Stage*)stg; // intptr_t back to pointer
stage->create_bullet(damage, "wowo"); // use pointer
}
Test_Class::Test_Class()
{
}
Test_Class::Test_Class(Test_Stage* stg, int num)
{
intptr_t stage = (intptr_t)stg;
// Create a new lua state
lua_State *myLuaState = luaL_newstate();
// Connect LuaBind to this lua state
luabind::open(myLuaState);
luaL_openlibs(myLuaState);

luabind::module(myLuaState)[
luabind::def("shoot_a_bullet", &shoot_a_bullet)//again this is static now
];
luabind::globals(myLuaState)["stage"] = stage;
cout << "I'm here " << endl;//just to check how far did the program go
luaL_dostring(
myLuaState,
"shoot_a_bullet(stage, 134)n"
);
cout << "I passed it"<< endl;
cin.get();
}
Test_Class::~Test_Class()
{
}    

玩得愉快！

我知道这很晚了，但你为什么不简单地这样做

luabind::module(myLuaState)[
luabind::class_<Test_Class>("Test_Class")
.def("shoot_a_bullet", &Test_Class::shoot_a_bullet)
];
luabind::globals(myLuaState)["stage"] = stage;
luaL_dostring(
myLuaState,
"stage::shoot_a_bullet(134)n"
);

你做的一切都对了，如果你想调用一个公开的C++对象的成员函数，你首先需要绑定它(你做到了)，然后你把你的对象放到某个lua作用域(你使用了全局)，然后你简单地引用对象阶段并使用：：myfunction调用一个函数