LuaL_openlibs（）和沙箱脚本 [英] LuaL_openlibs() and sandboxing scripts

问题描述

我在C / C ++应用程序中嵌入的Lua（5.1）。

I am embedding Lua (5.1) in a C/C++ application.

我使用的是 LuaL_openlibs（）函数加载基础库。但是，此功能加载，我要禁用，使他们不向我的Lua脚本一些其他库。

I am using the LuaL_openlibs() function to load the base libs. However, this function loads some other libraries which I want to disable so that they are not available to my Lua scripts.

具体来说，我要禁用IO和操作系统模块。是否有一个功能，我可以打电话programmativally禁用（或卸载），这些模块，这样我可以运行Lua脚本创造一个安全沙盒环境？

Specifically, I want to disable the IO and OS modules. Is there a function I can call to programmativally disable (or unload) these modules so that I can create a safe sandbox environment for running Lua scripts?

推荐答案

我不知道如何禁用模块，但你仍然可以选择加载，而不是装载它们全部用 luaL_openlibs <哪些/ code>。 Lua的5.1手动的第7.3节说：

I don't know how to disable modules, but you can still choose which ones to load instead of loading them all with luaL_openlibs. Section 7.3 of the Lua 5.1 manual says:

的 luaopen _ * 函数（打开库）不能直接调用，就像一个普通的C函数。他们必须通过Lua中被调用，就像一个Lua功能。

The luaopen_* functions (to open libraries) cannot be called directly, like a regular C function. They must be called through Lua, like a Lua function.

这是，而不是直接调用该函数在Lua 5.0：

That is, instead of directly calling the function as in Lua 5.0:

luaopen_table(L);

...你把它作为其名称的C函数，并使用 lua_call 或在Lua 5.1相似的：

... you push it as a C function with its name and use lua_call or similar in Lua 5.1:

lua_pushcfunction(L, luaopen_table);
lua_pushliteral(L, LUA_TABLIBNAME);
lua_call(L, 1, 0);

您可以在列出做到这一点的功能 lualib.h ：

The functions you can do this with are listed in lualib.h:

Function        | Name
----------------+-----------------
luaopen_base    | ""
luaopen_table   | LUA_TABLIBNAME
luaopen_io      | LUA_IOLIBNAME
luaopen_os      | LUA_OSLIBNAME
luaopen_string  | LUA_STRLIBNAME
luaopen_math    | LUA_MATHLIBNAME
luaopen_debug   | LUA_DBLIBNAME
luaopen_package | LUA_LOADLIBNAME

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