Thread.join阻塞主线程 [英] Thread.join blocks the main thread

问题描述

调用Thread.join会阻止当前(主)线程.但是，不调用join会导致在主线程退出时杀死所有产生的线程.如何在Ruby中生成持久子线程而不阻塞主线程?

Calling Thread.join blocks the current (main) thread. However not calling join results in all spawned threads to be killed when the main thread exits. How does one spawn persistent children threads in Ruby without blocking the main thread?

这是join的典型用法.

Here's a typical use of join.

for i in 1..100 do
  puts "Creating thread #{i}"
  t = Thread.new(i) do |j|
    sleep 1
    puts "Thread #{j} done"
  end
  t.join
end
puts "#{Thread.list.size} threads"

这给出了

     Creating thread 1  
     Thread 1 done  
     Creating thread 2  
     Thread 2 done  
     ...  
     1 threads  

但是我正在寻找如何获得这个

but I'm looking for how to get this

    Creating thread 1  
    Creating thread 2  
    ...  
    101 threads  
    Thread 1 done  
    Thread 2 done  
    ...  

代码在Ruby 1.8.7和1.9.2中提供相同的输出

The code gives the same output in both Ruby 1.8.7 and 1.9.2

推荐答案

您只需将线程堆积在另一个容器中，然后在创建完所有线程之后将它们逐个join:

You simply accumulate the threads in another container, then join them one-by-one after they've all been created:

my_threads = []
for i in 1..100 do
    puts "Creating thread #{i}"
    my_threads << Thread.new(i) do |j|
        sleep 1
        puts "Thread #{j} done"
    end
end
puts "#{Thread.list.size} threads"

my_threads.each do |t|
    t.join
end

您也不能将线程绑定到i变量，因为i会不断被覆盖，并且您的输出将是100行"Thread 100 done"(完成线程100)；相反，您必须将其绑定到i的副本，我巧妙地将其命名为j.

You also can't bind the thread to the i variable because i gets constantly overwritten, and your output will be 100 lines of "Thread 100 done"; instead, you have to bind it to a copy of i, which I have cleverly named j.

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