是否有内置方法来计算等待互斥锁的线程数?
例如:
m= Mutex.new
2.times do
Thread.new do
m.lock
end
end
m.nb_waiting_threads # => 1
答案 0 :(得分:6)
没有内置方式来计算等待Mutex
的线程,但是如果您可以将问题转换为使用Queue
,则会出现{{3}方法。
要使用Mutex
模拟Queue
,您将获得pop
的锁定,并通过push
值释放锁定。你的不变量是队列在任何给定时刻只包含0或1个项目。
require 'thread'
semaphore = Queue.new
semaphore.push(1) # Make synchronization token available
threads = []
5.times do |i|
threads << Thread.new do
semaphore.pop # Blocks until token available
puts "Thread #{i} working, #{semaphore.num_waiting} threads waiting."
sleep rand(3) # Do work
semaphore.push(1) # Release token
end
end
threads.each(&:join)
$ ruby queue_lock.rb
Thread 0 working, 0 threads waiting.
Thread 1 working, 3 threads waiting.
Thread 3 working, 2 threads waiting.
Thread 2 working, 1 threads waiting.
Thread 4 working, 0 threads waiting.