错误写法

在lua的userdata中分配内存

使用c++的placement new 调用 Mat 类的构造函数，把对象分配在lua的userdata空间中

Mat的生命周期由lua虚拟机管理，跟随lua变量生命周期

void createMat(const std::string& filename,Mat *result) {Mat img = cv::imread(filename, cv::IMREAD_COLOR);Mat* mat2 = new(result) Mat();*mat2 = img;
}

//在lua中申请内存
Mat* luaImage = static_cast<Mat*>(lua_newuserdata(L, sizeof(Mat)));
createMat(filename, luaImage);

这个写法不能调用Mat的析构函数会导致内存泄漏

正确写法

把Mat的析构函数注入到lua表里，名字为__gc

#include <opencv2/opencv.hpp>
#include <lua.hpp>using namespace cv;// GC function to be called by Lua
static int auto_gc(lua_State* L) {Mat* mat = *static_cast<Mat**>(lua_touserdata(L, 1));delete mat;return 0;
}void createMat(const std::string& filename, Mat** result) {Mat* img = new Mat(cv::imread(filename, cv::IMREAD_COLOR));*result = img;
}extern "C" int createMat_lua(lua_State* L) {const char* filename = luaL_checkstring(L, 1);// 创建一个新的tablelua_newtable(L);// 创建userdata保存Mat*指针Mat** luaImage = static_cast<Mat**>(lua_newuserdata(L, sizeof(Mat*)));createMat(filename, luaImage);// 将userdata放入table中lua_setfield(L, -2, "mat");// 创建元表if (luaL_newmetatable(L, "MatMetaTable")) {lua_pushcfunction(L, auto_gc);lua_setfield(L, -2, "__gc");}// 将元表设置为table的元表lua_setmetatable(L, -2);return 1; // 返回table
}extern "C" int luaopen_my_module(lua_State* L) {lua_register(L, "createMat", createMat_lua);return 1;
}int main() {lua_State* L = luaL_newstate();luaL_openlibs(L);luaopen_my_module(L);// Run Lua scriptluaL_dofile(L, "script.lua");lua_close(L);return 0;
}