Ruby 中的陷阱处理程序是否存在并发限制？

Question

我运行以下代码，（一个或多个）CLD 陷阱丢失，从而留下一个已失效（僵尸）进程，该进程尚未使用 Process.wait 收集其退出状态

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所需数量的并发子出口？

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

谢谢...

Answer 1

我已经在 Python 中解决了类似的问题。根据 ruby​​ 实现其信号处理的方式，如果子进程在陷阱处理程序的先前调用正在运行时终止，则陷阱处理程序可能不会再次被调用。

因此，安全的设计是使用循环并在单个陷阱处理程序运行中获取所有可能的子级。要做到这一点，假设您已经有了孩子的名单 - 这就是您的情况。

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

Answer 2

惯用的 Ruby 提示：

1. 多行时，4.times {|i| 通常是 4.times do |i|。
2. pp Children 可以替换为 "waiting for #{children.inspect}" 除非您希望将这两件事放在不同的行上。