从.lua的使用句柄调用lua函数？ [英] Calling lua functions from .lua's using handles?
问题描述
我正在开发一个小项目，试图将lua与c ++集成。
然而我的问题如下：
我有多个lua脚本，让他们称为s1.lua s2.lua和s3.lua。每个都有以下函数：setVars（）和executeResults（）。
现在我可以通过LuaL_dofile调用lua文件，和/或executeResults（）。这里的问题是，我加载s2.lua后，我不能再调用s1.lua的函数。这意味着我必须在s1.lua上重做LuaL_dofile以重新获得对函数的访问权限，这样我就无法访问s2.lua中的函数。
有一种方法来简单地加载所有lua文件在一行，然后开始调用他们的功能随意？像s1-> executeResults（）s5-> executeResults（）s3-> setVars（）等。
我现在已经有一个系统使用boost :: filesystem检测文件夹中的所有lua文件，然后将这些文件名保存在向量中，然后简单地遍历该向量，以便在一行中加载每个lua文件。
前进用Lua文件名填充向量我的插件加载函数看起来像这样：
void Lua_plugin :: load_Plugins ）{
std :: vector< std :: string> :: const_iterator it;
for（it = Lua_PluginList.begin（）; it！= Lua_PluginList.end（）; it ++）{
std :: cout<
std :: string filepath = scriptdir + filename;
if（luaL_loadfile（L，filepath.c_str（））|| lua_pcall（L，0，0，0））{
std :: cout< ScriptEngine：错误加载脚本。返回的错误是：< lua_tostring（L，-1）< std :: endl;
}
}
}
清除，所有我在.lua的是这样的：
- s1.lua
setVars（）
--do stuff
end
executeResults（）
--dostuff
end
等等，但我想能够调用s1.lua的setVars（）和s2.lua的setVars
这实际上是使用C API建议的 b
$ b
#include< stdio.h>
#includelua.h
static void
executioncript（lua_State * L，const char * filename，const char * function）
{
/ *从区域中检索环境* /
lua_getfield（L，LUA_REGISTRYINDEX，filename）;
/ *从环境中获取所需的函数* /
lua_getfield（L，-1，function）;
return lua_call（L，0，0）;
}
static void
loadscript（lua_State * L，const char * filename）
{
/ *将lua脚本加载到内存中* /
luaL_loadfile（L，filename）;
/ *创建一个新的函数环境并将其存储在注册表中* /
lua_createtable（L，0，1）;
lua_getglobal（L，print）;
lua_setfield（L，-2，print）;
lua_pushvalue（L，-1）;
lua_setfield（L，LUA_REGISTRYINDEX，filename）;
/ *设置加载脚本的环境并执行它* /
lua_setfenv（L，-2）;
lua_call（L，0，0）;
/ *运行脚本初始化函数* /
executioncript（L，filename，init）;
}
int
main（int argc，char * argv []）
{
lua_State * L;
int env1，env2;
L =（lua_State *）luaL_newstate（）;
luaL_openlibs（L）;
loadscript（L，test1.lua）;
loadscript（L，test2.lua）;
executioncript（L，test1.lua，run）;
executioncript（L，test2.lua，run）;
executioncript（L，test2.lua，run）;
executioncript（L，test1.lua，run）;
return 0;
}
测试脚本：
- test1.lua
function init（）output ='test1'end
function run（）print（output）end
$ b b - test2.lua
function init（）output ='test2'end
function run（）print（output）end
输出：
test1
test2
test2
test1
$ b $ p
为了简洁，省略了所有的错误处理，但是你需要检查返回值 luaL_loadfile 并使用 lua_pcall 而不是 lua_call / p>
I'm working on a small project trying to integrate lua with c++. My problem however is as follows:
I have multiple lua scripts, lets call them s1.lua s2.lua and s3.lua. Each of these has the following functions: setVars() and executeResults().
Now I am able to to call a lua file through LuaL_dofile and immediately after use setVars() and/or executeResults(). The problem here however is that after I load s2.lua I can no longer call the functions of s1.lua. This would mean I have to redo the LuaL_dofile on s1.lua to regain access to the function and by doing so I lose access to the functions in s2.lua.
Is there a way to simply load all lua files in a row, and afterwards start calling their functions at will? Something like s1->executeResults() s5->executeResults() s3->setVars() etc.
I currently already have a system in place using boost::filesystem to detect all lua files in a folder, I then save these files names in a vector and then simply iterate over the vector to load each lua file in a row.
Foregoing the filling of the vector with lua file names my plugin load function looks like this at the moment:
void Lua_plugin::load_Plugins(){
std::vector<std::string>::const_iterator it;
for (it=Lua_PluginList.begin(); it!=Lua_PluginList.end(); it++){
std::cout<<"File loading: " << *it << std::endl;
std::string filename = *it;
std::string filepath = scriptdir+filename;
if (luaL_loadfile(L, filepath.c_str()) || lua_pcall(L, 0, 0, 0)) {
std::cout << "ScriptEngine: error loading script. Error returned was: " << lua_tostring(L, -1) << std::endl;
}
}
}
To make it a bit more clear, all I have in the .lua's is something like this:
-- s1.lua
setVars()
--do stuff
end
executeResults()
--dostuff
end
etc, but I would like to be able to call s1.lua's setVars() and s2.lua's setVars() after simply having loaded both in a row.
This is effectively what gwell proposed using the C API:
#include <stdio.h>
#include "lua.h"
static void
executescript(lua_State *L, const char *filename, const char *function)
{
/* retrieve the environment from the resgistry */
lua_getfield(L, LUA_REGISTRYINDEX, filename);
/* get the desired function from the environment */
lua_getfield(L, -1, function);
return lua_call(L, 0, 0);
}
static void
loadscript(lua_State *L, const char *filename)
{
/* load the lua script into memory */
luaL_loadfile(L, filename);
/* create a new function environment and store it in the registry */
lua_createtable(L, 0, 1);
lua_getglobal(L, "print");
lua_setfield(L, -2, "print");
lua_pushvalue(L, -1);
lua_setfield(L, LUA_REGISTRYINDEX, filename);
/* set the environment for the loaded script and execute it */
lua_setfenv(L, -2);
lua_call(L, 0, 0);
/* run the script initialization function */
executescript(L, filename, "init");
}
int
main(int argc, char *argv[])
{
lua_State *L;
int env1, env2;
L = (lua_State *) luaL_newstate();
luaL_openlibs(L);
loadscript(L, "test1.lua");
loadscript(L, "test2.lua");
executescript(L, "test1.lua", "run");
executescript(L, "test2.lua", "run");
executescript(L, "test2.lua", "run");
executescript(L, "test1.lua", "run");
return 0;
}
Test scripts:
-- test1.lua
function init() output = 'test1' end
function run() print(output) end
-- test2.lua
function init() output = 'test2' end
function run() print(output) end
Output:
test1
test2
test2
test1
I omitted all error handling for brevity, but you'll want to check the return value of luaL_loadfile and use lua_pcall instead of lua_call.
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