如何执行Lua脚本的运行时限制? [英] How to enforce Lua scripts runtime limit?

问题描述

在doa中使用dofile在lua中运行一个几乎平凡的脚本，运行10000次，大约需要52秒，但是如果我运行10000次"lua52 script.lua"，则需要运行3到4倍.我知道有更多的系统调用和其他开销，但是我尝试实现的是运行脚本，超时时间为3秒，然后输出输出.我的问题是有无限循环的脚本，有意或无意，例如:

Running an almost trivial script in lua with dofile, 10000 times, takes about 52 seconds in this machine, but if i run 10000 times "lua52 script.lua", it takes 3 or 4 times more. I'm aware that there's more system calls involved and other overhead, but what i try to achieve is running scripts with a timeout of let's say 3 seconds, and print out the output. My problem is scripts with infinite loops, intentional or not, for example:

while(true) do
end

我可以在Lua内为dofile超时吗?我唯一的选择是每次都用timeout(3)调用解释器吗?

Can i make a timeout for a dofile from within Lua? Is my only option to call the interpreter each time with timeout(3)?

推荐答案

对于像我这样的新手来说，在Lua问题上更正lhf感觉有点不对劲，但这是可行的.将"count"传递给debug.sethook与传递"c"或"call"相同，在 n VM指令为"之后传递用于触发关联函数的正确掩码.

It feels a bit wrong for a novice like me to be correcting lhf on Lua matters, but here goes; passing "count" to debug.sethook is the same as passing "c" or "call", the correct mask to pass to fire the associated function after n VM instructions is "".

因此，要限制从dofile()加载的代码的运行时间，请使用以下内容:

As such, to restrict the runtime of code loaded from dofile(), use something like the following:

local f = function() error("timeout") end
local x,y = xpcall(function()
  debug.sethook(f, "", 1e8)
  local ret = dofile("script.lua")
  debug.sethook()
  return ret
end, debug.traceback)

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