Question

我正在阅读Jesse Storimer的优秀书籍Working with Unix Processes。在有关从已退出的子进程捕获信号的部分中，他给出了一个代码示例。

我稍微修改了这段代码（见下文），以便更清楚地了解正在发生的事情：

父母在信号之间恢复自己的执行（我可以用puts看到），
wait在一个trap声明中为多个孩子执行{有时我收到“收到一个CHLD信号”，然后是多个“子pid退出”）。

预期产出

通常，以下代码的输出类似于：

parent is working hard
Received a CHLD signal
child pid 73408 exited
parent is working hard
parent is working hard
parent is working hard
Received a CHLD signal
child pid 73410 exited
child pid 73409 exited
All children exited - parent exiting too.

偶然的错误

但偶尔我会收到这样的错误：

trapping_signals.rb:17:in `write': deadlock; recursive locking (ThreadError)
    from trapping_signals.rb:17:in `puts'
    from trapping_signals.rb:17:in `puts'
    from trapping_signals.rb:17:in `block in <main>'
    from trapping_signals.rb:17:in `call'
    from trapping_signals.rb:17:in `write'
    from trapping_signals.rb:17:in `puts'
    from trapping_signals.rb:17:in `puts'
    from trapping_signals.rb:17:in `block in <main>'
    from trapping_signals.rb:40:in `call'
    from trapping_signals.rb:40:in `sleep'
    from trapping_signals.rb:40:in `block in <main>'
    from trapping_signals.rb:38:in `loop'
    from trapping_signals.rb:38:in `<main>

有谁可以向我解释这里出了什么问题？

代码

child_processes = 3
dead_processes = 0

# We fork 3 child processes.
child_processes.times do
  fork do
    # Each sleeps between 0 and 5 seconds
    sleep rand(5)
  end
end

# Our parent process will be busy doing some work.
# But still wants to know when one of its children exits.

# By trapping the :CHLD signal our process will be notified by the kernel
# when one of its children exits.
trap(:CHLD) do
  puts "Received a CHLD signal"
  # Since Process.wait queues up any data that it has for us we can ask for it
  # here, since we know that one of our child processes has exited.

  # We loop over a non-blocking Process.wait to ensure that any dead child
  # processes are accounted for.
  # Here we wait without blocking.
  while pid = Process.wait(-1, Process::WNOHANG)
    puts "child pid #{pid} exited"
    dead_processes += 1

    # We exit ourselves once all the child processes are accounted for.
    if dead_processes == child_processes
      puts "All children exited - parent exiting too."
      exit
    end
  end
end

# Work it.
loop do
  puts "parent is working hard"
  sleep 1
end

Answer 1

我浏览了Ruby sources以查看引发该特定错误的位置，并且只在当前线程尝试获取锁定时才会引发该错误，但当前线程已经采用了相同的锁定。这意味着锁定不是可重入的：

m = Mutex.new
m.lock
m.lock #=> same error as yours

现在至少我们知道会发生什么，但不知道为什么以及在哪里。错误消息表明它在调用puts期间发生。当它被调用时，它最终会在io_binwrite中结束。 stdout未同步，但它是缓冲的，因此在第一次调用时满足this if condition，并且将设置缓冲区加上该缓冲区的写锁定。写锁定对于保证写入stdout的原子性非常重要，不应该发生同时写入stdout的两个线程混淆彼此的输出。为了证明我的意思：

t1 = Thread.new { 100.times { print "aaaaa" } }
t2 = Thread.new { 100.times { print "bbbbb" } }
t1.join
t2.join

尽管两个线程都在写入stdout，但是一次写入就不会发生 - 你将始终按顺序排列完整的5个或b个。这就是写锁is there for。

现在你的情况出了什么问题就是写锁的竞争条件。父进程每秒循环并写入stdout（“父进程正在努力”）。但是同一个线程最终也会执行trap块并再次尝试写入stdout（“收到CHLD信号”）。您可以通过在#{Thread.current}语句中添加puts来验证它是否真的是同一个线程。如果这两个事件发生得足够紧密，那么你将遇到与第一个例子相同的情况：同一个线程尝试两次获得相同的锁，这最终会触发错误。

是什么导致我的Ruby`track`块中出现这种死锁？

预期产出

偶然的错误

代码

1 个答案: