如何计算线程等待互斥量？

Question

是否有内置的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

Answer 1

没有内置的方法来计算等待a的线程Mutex,但是如果你可以将你的问题转换为使用a Queue,那么就有一种num_waiting方法.

要Mutex使用a 模拟a Queue,您将获得锁定pop并通过push值释放锁定.您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.