在C++中，使用luabind，调用lua文件中定义的函数？

发布于 2024-11-14 23:31:15 字数 771 浏览 5 评论 0原文

假设我有一个 lua 文件：

--functions.lua
function testadd(a, b) 
    return a+b
end

我将如何使用 luabind 加载该文件，并调用该函数 - 类似于：

//test.cpp
extern "C" {
    #include "lua.h"
    #include "lualib.h"
    #include "lauxlib.h"
}
#include <luabind/luabind.hpp>
#include <luabind/function.hpp> 

int main() {
    lua_State *myLuaState = lua_open();
    luaL_openlibs(myLuaState);
    luaL_loadfile(myLuaState, "functions.lua");
    luabind::open(myLuaState);
    int value = luabind::call_function<int>(myLuaState, "testadd", 2, 3);
    lua_close(myLuaState);
}

但这会返回一个错误： 抛出“luabind::error”实例后调用终止 What(): lua运行时错误中止

那么，做我想做的事情的正确语法是什么？（从错误的外观看来，这似乎是lua文件中语法的问题，但我不认为这是......）

原文

Say I have a lua file:

--functions.lua
function testadd(a, b) 
    return a+b
end

How would I use luabind to load that file, and call that function- something like:

//test.cpp
extern "C" {
    #include "lua.h"
    #include "lualib.h"
    #include "lauxlib.h"
}
#include <luabind/luabind.hpp>
#include <luabind/function.hpp> 

int main() {
    lua_State *myLuaState = lua_open();
    luaL_openlibs(myLuaState);
    luaL_loadfile(myLuaState, "functions.lua");
    luabind::open(myLuaState);
    int value = luabind::call_function<int>(myLuaState, "testadd", 2, 3);
    lua_close(myLuaState);
}

But this returns an error:
terminate called after throwing an instance of 'luabind::error'
what(): lua runtime error
Aborted

So, what is the proper syntax for doing what I want to do?
(From the looks of the error it seems to be a problem with the syntax in the lua file, but I don't think it is...)

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