Rom*_*ian 3 ruby multithreading synchronization mutex monitor
我的问题是我不知道如何使用Ruby同步多个线程。该任务是创建六个线程并立即启动它们。他们都应该puts "Thread 1" Hi"按我需要的顺序依次进行一些工作(例如)。
经过一段时间与Mutex和Condition Variable的斗争后,我实现了我的目标。这段代码有点混乱,我故意不使用循环来获得“更清晰的视图”。
cv = ConditionVariable.new
mutex = Mutex.new
mutex2 = Mutex.new
cv2 = ConditionVariable.new
mutex3 = Mutex.new
cv3 = ConditionVariable.new
mutex4 = Mutex.new
cv4 = ConditionVariable.new
mutex5 = Mutex.new
cv5 = ConditionVariable.new
mutex6 = Mutex.new
cv6 = ConditionVariable.new
Thread.new do
mutex.synchronize {
puts 'First: Hi'
cv.wait(mutex)
puts 'First: Bye'
#cv.wait(mutex)
cv.signal
puts 'First: One more time'
}
end
Thread.new do
mutex.synchronize {
puts 'Second: Hi'
cv.signal
cv.wait(mutex)
puts 'Second:Bye'
cv.signal
}
mutex2.synchronize {
puts 'Second: Starting third'
cv2.signal
}
end
Thread.new do
mutex2.synchronize {
cv2.wait(mutex2)
puts 'Third: Hi'
}
mutex3.synchronize {
puts 'Third: Starting forth'
cv3.signal
}
end
Thread.new do
mutex3.synchronize {
cv3.wait(mutex3)
puts 'Forth: Hi'
}
mutex4.synchronize {
puts 'Forth: Starting fifth'
cv4.signal
}
end
Thread.new do
mutex4.synchronize {
cv4.wait(mutex4)
puts 'Fifth: Hi'
}
mutex5.synchronize {
puts 'Fifth: Starting sixth'
cv5.signal
}
end
Thread.new {
mutex5.synchronize {
cv5.wait(mutex5)
puts 'Sixth:Hi'
}
}
sleep 2
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您可以滥用Queue,将其像传统的PV信号灯一样使用。为此,您创建一个Queue实例:
require 'thread'
...
sem = Queue.new
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当线程需要等待时,它将调用Queue#deq:
# waiting thread
sem.deq
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当其他某个线程想要解除阻塞正在等待的线程时,它将某些东西(任何东西)推送到队列中:
# another thread that wants to unblock the waiting thread
sem.enq :go
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这是一个使用Queue同步其开始和停止的工作程序类:
class Worker
def initialize(worker_number)
@start = Queue.new
Thread.new do
@start.deq
puts "Thread #{worker_number}"
@when_done.call
end
end
def start
@start.enq :start
end
def when_done(&block)
@when_done = block
end
end
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构造后,工作程序会创建一个线程,但是该线程然后在@start队列中等待。直到调用#start,线程才会解除阻塞。
完成后,线程将执行已调用#when_done的块。我们将在短时间内看到如何使用它。
首先,让我们确保如果有任何线程引发异常,我们将对其进行了解:
Thread.abort_on_exception = true
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我们将需要六名工人:
workers = (1..6).map { |i| Worker.new(i) }
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这是#when_done起作用的地方:
workers.each_cons(2) do |w1, w2|
w1.when_done { w2.start }
end
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这轮流带动每对工人。除最后一个工人外,每个工人都被告知,完成时应在其之后启动该工人。剩下的只有最后一个工人。完成后,我们希望它通知该线程:
all_done = Queue.new
workers.last.when_done { all_done.enq :done }
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现在剩下的就是启动第一个线程:
workers.first.start
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并等待最后一个线程完成:
all_done.deq
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输出:
Thread 1
Thread 2
Thread 3
Thread 4
Thread 5
Thread 6
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