ruby - Ruby 中的陷阱处理程序是否存在一些并发限制？

Question

我运行以下代码，(一个或多个)CLD 陷阱丢失，从而留下一个失效的(僵尸)进程，该进程尚未通过使用 Process.wait

require 'pp'

children = []
trap("CLD") do
  cpid = Process.wait
  puts "CLD from pid #{cpid} at #{Time.now}"
  6.times {|i| puts "  ... Waiting[#{i}] in CLD trap for pid #{cpid}"; sleep 0.5}
  puts "OK, finished slow trap at #{Time.now} for pid #{cpid}"
  children.delete cpid
end

4.times {|i|
    if child_pid = fork # parent
      puts "In parent, child_pid[#{i}] = #{child_pid}"
      children.push child_pid
    else
      puts "In child[#{i}], PID=#{$$}"
      sleep 0.2
      puts "In child[#{i}], PID=#{$$} ... exiting"
      exit!
    end
  }

while true
  sleep 2
  exit if children.length == 0
  puts "[#{Time.now}] ... Parent still waiting for \n"
  pp children
  sleep 8
end

这是一个示例运行输出:

[admin@jcmsa pe2]# ruby multitrap.rb
In parent, child_pid[0] = 9285
In child[0], PID=9285
In parent, child_pid[1] = 9289
In child[1], PID=9289
In parent, child_pid[2] = 9293
In child[2], PID=9293
In parent, child_pid[3] = 9297
In child[3], PID=9297
In child[0], PID=9285 ... exiting
In child[3], PID=9297 ... exiting
In child[1], PID=9289 ... exiting
In child[2], PID=9293 ... exiting
CLD from pid 9285 at 2011-02-03 13:31:20 -0800
  ... Waiting[0] in CLD trap for pid 9285
CLD from pid 9289 at 2011-02-03 13:31:20 -0800
  ... Waiting[0] in CLD trap for pid 9289
CLD from pid 9293 at 2011-02-03 13:31:20 -0800
  ... Waiting[0] in CLD trap for pid 9293
  ... Waiting[1] in CLD trap for pid 9293
  ... Waiting[2] in CLD trap for pid 9293
  ... Waiting[3] in CLD trap for pid 9293
  ... Waiting[4] in CLD trap for pid 9293
  ... Waiting[5] in CLD trap for pid 9293
OK, finished slow trap at 2011-02-03 13:31:23 -0800 for pid 9293
  ... Waiting[1] in CLD trap for pid 9289
  ... Waiting[2] in CLD trap for pid 9289
  ... Waiting[3] in CLD trap for pid 9289
  ... Waiting[4] in CLD trap for pid 9289
  ... Waiting[5] in CLD trap for pid 9289
OK, finished slow trap at 2011-02-03 13:31:25 -0800 for pid 9289
  ... Waiting[1] in CLD trap for pid 9285
  ... Waiting[2] in CLD trap for pid 9285
  ... Waiting[3] in CLD trap for pid 9285
  ... Waiting[4] in CLD trap for pid 9285
  ... Waiting[5] in CLD trap for pid 9285
OK, finished slow trap at 2011-02-03 13:31:28 -0800 for pid 9285
[2011-02-03 13:31:28 -0800] ... Parent still waiting for
[9297]
[2011-02-03 13:31:38 -0800] ... Parent still waiting for
[9297]
[2011-02-03 13:31:48 -0800] ... Parent still waiting for
[9297]
[2011-02-03 13:31:58 -0800] ... Parent still waiting for
[9297]

……等等……

然后 'ps axf' 显示

 9283 pts/2    Sl+    0:00  |   |       \_ ruby multitrap.rb
 9297 pts/2    Z+     0:00  |   |           \_ [ruby] <defunct>

在我的实验中

  3 children .... sometimes gets a zombie
  4 children .... more often gets a zombie
  5 children .... always gets a zombie, sometimes more than one

这里有什么限制？

如何设置 CLD 陷阱处理程序来处理我需要的尽可能多的并发子退出？

ruby 版本是 ruby 1.9.1p243 (2009-07-16 revision 24175) [x86_64-linux]

谢谢...

Answer 1

我已经解决了 Python 中的类似问题。根据 ruby​​ 如何实现其信号处理，如果子进程终止而陷阱处理程序的先前调用正在运行，则可能不会再次调用陷阱处理程序。

因此，一个安全的设计是使用一个循环，并在一次陷阱处理程序运行中收集所有可能的 child 。要做到这一点，假设您已经有了 child 的列表 - 这就是您的情况。

--陷阱( child ) 对于每个 child_pid in children rc = waitpid(pid, WNOHANG) 如果收获则从列表中删除条目