Question

我遇到了同步线程的问题，我不知道怎么做，有人能帮帮我吗？

所以，问题是我必须以某种特定的顺序启动线程。订单如下：

线程1和7可以同时进行，并且在其中一个完成之后，下一个线程被启动（即线程2或/和线程6），线程3和5的方式相同。最后一个，在线程3和5完成运行之后，进入最后一个，线程4。

这是我开始使用的代码，但我在某种程度上陷入了队列实现。

MUTEX          = Mutex.new
high_condition = ConditionVariable.new
low_condition  = ConditionVariable.new
threads = []

7.times do |i|
  threads << Thread.new{
    MUTEX.synchronize {
      Thread.current["number"] = i

      you_shall_not_pass
    }
  }
end

threads.map(&:join)

def you_shall_not_pass
  order = Thread.current["number"]
end

Answer 1

使用Ruby的Queue作为counting semaphore。它具有阻塞push和pop操作，您可以使用这些操作将有限数量的令牌分发给线程，要求每个线程在运行之前获取令牌并在令牌完成时释放令牌。如果使用2个令牌初始化队列，则可以确保一次只运行2个线程，并且可以按照自己喜欢的顺序创建线程。

require 'thread'

semaphore = Queue.new
2.times { semaphore.push(1) } # Add two concurrency tokens

puts "#{semaphore.size} available tokens"

threads = []
[1, 7, 2, 6, 3, 5, 4].each do |i|
  puts "Enqueueing thread #{i}"
  threads << Thread.new do
    semaphore.pop # Acquire token
    puts "#{Time.now} Thread #{i} running. #{semaphore.size} available tokens. #{semaphore.num_waiting} threads waiting."
    sleep(rand(10)) # Simulate work
    semaphore.push(1) # Release token
  end
end

threads.each(&:join)

puts "#{semaphore.size} available tokens"

$ ruby counting_semaphore.rb
2 available tokens
Enqueueing thread 1
Enqueueing thread 7
2015-12-04 08:17:11 -0800 Thread 7 running. 1 available tokens. 0 threads waiting.
2015-12-04 08:17:11 -0800 Thread 1 running. 0 available tokens. 0 threads waiting.
Enqueueing thread 2
Enqueueing thread 6
2015-12-04 08:17:11 -0800 Thread 2 running. 0 available tokens. 0 threads waiting.
Enqueueing thread 3
Enqueueing thread 5
Enqueueing thread 4
2015-12-04 08:17:19 -0800 Thread 6 running. 0 available tokens. 3 threads waiting.
2015-12-04 08:17:19 -0800 Thread 5 running. 0 available tokens. 2 threads waiting.
2015-12-04 08:17:21 -0800 Thread 3 running. 0 available tokens. 1 threads waiting.
2015-12-04 08:17:22 -0800 Thread 4 running. 0 available tokens. 0 threads waiting.
2 available tokens

Ruby多线程队列

1 个答案: