我试图userdata
从 C++ 中的 Lua 脚本(chunk A
)获取(通过我的示例中的函数返回的变量),然后,稍后将其从 C++(通过我的示例中的函数参数)传递userdata
回 Lua 脚本(chunk B
),所以userdata
可以chunk B
像在chunk A
.
MyBindings.h
class Vec2
{
public:
Vec2():x(0), y(0){};
Vec2(float x, float y):x(x), y(y){};
float x, y;
};
MyBindings.i
%module my
%{
#include "MyBindings.h"
%}
%include "MyBindings.h"
main.cpp
#include <iostream>
#include <lua.hpp>
extern "C"
{
int luaopen_my(lua_State *L);
}
int main()
{
lua_State *L = luaL_newstate();
luaL_openlibs(L);
luaopen_my(L);
lua_settop(L, 0);
/* chunk A */
luaL_dostring(L, "local vec2 = my.Vec2(3, 4)\n"
"function setup()\n"
"return vec2\n"
"end\n");
/* chunk B */
luaL_dostring(L, "function test(p)\n"
"print(p.x)\n"
"end\n");
void *userDataPtr = nullptr;
/* call setup function */
int top = lua_gettop(L);
lua_getglobal(L, "setup");
if (lua_pcall(L, 0, LUA_MULTRET, 0))
{
std::cout << lua_tostring(L, -1) << '\n';
lua_pop(L, 1);
}
/* check the return value */
if (lua_gettop(L) - top)
{
/* store userdata to a pointer */
if (lua_isuserdata(L, -1))
userDataPtr = lua_touserdata(L, -1);
}
/* check if userDataPtr is valid */
if (userDataPtr != nullptr)
{
/* call test function */
lua_getglobal(L, "test");
lua_pushlightuserdata(L, userDataPtr); /* pass userdata as an argument */
if (lua_pcall(L, 1, 0, 0))
{
std::cout << lua_tostring(L, -1) << '\n';
lua_pop(L, 1);
}
}
lua_close(L);
}
我得到的结果:
[string "local vec2 = my.Vec2(3, 4)..."]:6: 尝试索引用户数据值 (local 'p')
我期望的结果:
3
是否可以从中获取userdata
然后chunk A
将其传递给chunk B
以便可以像以前一样使用它chunk A
?