防止Lua的无限循环 [英] Prevent Lua infinite loop

问题描述

我用LUA接口来获得LUA支持我的C＃程序，如果用户提交这样的代码工作线程将冻结

I use lua interfaces to get lua support in my C# program, the worker thread will freeze if the user submits code like this

while true do end

我有一个办法，如果一个无限循环运行检测，但我需要在退出从工作线程DoString方法的一个很好的方式。 ？任何想法

I have a way to detect if a infinite loop is running, but I need a nice way of exiting the DoString method from the Worker thread. Any ideas?

编辑：@kikito，是林deteting它类似的东西。我的问题是，我无法找到杀死DoString方法的干净的方式，它看起来像Lua的接口主类（LUA）有一些静态依赖，因为如果我这样做 lua.Close（）; 在我的实例，它将中止DoString方法，但下一次我实例的LUA类新的Lua（）; 就会死机说着有关保护内存

edit: @kikito, Yes Im deteting it something like that. The problem I have is that I cant find a clean way of killing the DoString method, it looks like the Lua interfaces main class (Lua) has some static dependencies, because if I do lua.Close(); on my instance it will abort the DoString method, but the next time I instance an lua class new Lua(); it will crash saying something about protective memory

编辑：一个特性分支显示我的.Close代码
的 https://github.com/AndersMalmgren/FreePIE/tree/detect-and-recover-infite-lua-loop

edit: A feature branch showing my .Close code https://github.com/AndersMalmgren/FreePIE/tree/detect-and-recover-infite-lua-loop

推荐答案

设置挂钩是不充分的所有，以防止资源浪费意外，更不用说abuse-这里有一个简单的例子（字符串模式中的时间都花在匹配 - 没有调试钩子被调用）：

Sandboxing Lua

Setting hooks is not sufficient at all to prevent unintended waste of resources, let alone abuse- here's a simple example (the time is spent during string pattern matching- no debug hooks get called):

S =（'A'）：代表（20000）：比赛（'.- b'）

< STRONG>的唯一可靠方法来强制在一块Lua代码时间/内存限制是在其自己的进程中运行Lua解释，使您的OS监控过程。

和Lua的好处是，你不会需要任何沙箱复杂，依赖于操作系统的权限设定：您刚才限制时间和内存（合理;在Windows上有招聘对象，Unix有相关ulimits- ： Linux的资源限​​制），然后让事情像os.execute，在IO库，并从编剧（很容易）的距离就像luasocket模块的一半。

The nice thing with Lua is that you won't need any complicated, OS-dependent permission setup for sandboxing: you just limit time and memory (reasonable; on windows there are Job Objects, Unix has ulimits- relevant: Linux resource limitation) and then keep things like os.execute, half the io-library and modules like luasocket away from the scripter (pretty easy).

您可以在不捣毁你的Lua解释处理几乎一切（除了违反了时间/内存限制）：只是包装用户提供的代码的pcall ;如果你调用可能会失败，任何个人的Lua-API函数，你需要用他们，你可以PCALL，太（或设置一个Lua恐慌功能，并从那里处理它）函数内。

You can handle almost everything (except violation of time/memory limits) without trashing your Lua interpreter: Just wrap the execution of user-supplied code inside a pcall; if you call any Lua-API functions that might fail yourself, you need to wrap them inside a function that you can pcall, too (or set a Lua panic function and handle it from there).

[我不想让人们在这个线程一眼假设debug.sethook是足够的沙盒和计算器不会让我评论（还）]

[I didn't want people glancing at this thread to assume that debug.sethook is adequate for sandboxing, and stackoverflow would not let me comment (yet)]

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