在代码中最容易解释:
- 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
我的两个模块的心理模型是一样的:
那么,我失踪了什么?
编辑:drmaciver在twitter上建议,在第一种情况下,由于某些原因,管道插座进入非阻塞模式,但在第二种情况下.我不能想到为什么会发生这种情况,也不能弄清楚如何获得描述符的标志,但至少是一个合理的答案?努力解决这种可能性.
解决方法
阿哈,微妙.
在第二种情况下,在IO#popen块的末尾有一个隐藏的,阻塞的保证子句. Timeout :: Error被及时提升,但是在执行从该隐式确认子句返回之前,您无法挽救它.
Under the hood,IO.popen(cmd){| io | …}这样做:
- 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
而IO#close调用实际上是一个或多或少的pclose(3),它阻止你进入waitpid(2)直到睡着的孩子退出.
你可以这样验证:
- #!/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?
您必须以显式的方式执行此操作,因为解释器不会公开一种方法来覆盖IO#popen确认逻辑.您可以使用上述代码作为起始模板,并取消注释kill()行,例如.