popen中的超时工作,但是超时内的popen不行吗？

Question

在代码中解释最简单:

require 'timeout'

puts "this block will properly kill the sleep after a second"

IO.popen("sleep 60") do |io|
  begin
    Timeout.timeout(1) do
      while (line=io.gets) do
        output += line
      end
    end
  rescue Timeout::Error => ex
    Process.kill 9, io.pid
    puts "timed out: this block worked correctly"
  end
end

puts "but this one blocks for >1 minute"

begin
  pid = 0
  Timeout.timeout(1) do
    IO.popen("sleep 60") do |io|
      pid = io.pid
      while (line=io.gets) do
        output += line
      end
    end
  end
rescue Timeout::Error => ex
  puts "timed out: the exception gets thrown, but much too late"
end

我的两个街区的心理模型是相同的:

那么,我错过了什么？

编辑:drmaciver在twitter上建议,在第一种情况下,由于某种原因,管道套接字进入非阻塞模式,但在第二种情况下它不会.我想不出有什么理由会发生这种情况,也无法弄清楚如何获得描述符的标志,但它至少是一个看似合理的答案？研究这种可能性.

Answer 1

啊哈,微妙.

ensure在第二种情况下,在IO#popen块的末尾有一个隐藏的阻塞子句.Timeout :: Error 是及时引发的,但是rescue直到执行从该隐式ensure子句返回时才能执行.

在幕后,IO.popen(cmd) { |io| ... }做这样的事情:

def my_illustrative_io_popen(cmd, &block)
  begin
    pio = IO.popen(cmd)
    block.call(pio)      # This *is* interrupted...
  ensure
    pio.close            # ...but then control goes here, which blocks on cmd's termination
  end

并且IO#close调用实际上是或多或少pclose(3),这会阻止你进入,waitpid(2)直到睡着的孩子退出.

您可以这样验证:

#!/usr/bin/env ruby

require 'timeout'

BEGIN { $BASETIME = Time.now.to_i }

def xputs(msg)
  puts "%4.2f: %s" % [(Time.now.to_f - $BASETIME), msg]
end

begin
  Timeout.timeout(3) do
    begin
      xputs "popen(sleep 10)"
      pio = IO.popen("sleep 10")
      sleep 100                     # or loop over pio.gets or whatever
    ensure
      xputs "Entering ensure block"
      #Process.kill 9, pio.pid      # <--- This would solve your problem!
      pio.close
      xputs "Leaving ensure block"
    end
  end
rescue Timeout::Error => ex
  xputs "rescuing: #{ex}"
end

所以,你可以做什么？

您必须以显式方式执行此操作,因为解释器不会公开覆盖IO#popen ensure逻辑的方法.例如,您可以使用上面的代码作为起始模板并取消注释该kill()行.