Ruby：等待使用join和ThreadsWait.all_waits完成的所有线程 –

Question

请考虑以下示例：

threads = []

(0..10).each do |_|
  threads << Thread.new do
    # do async staff there
    sleep Random.rand(10)
  end
end

然后有2种方法可以在完成时等待：

>使用join：

threads.each(&:join)

>使用ThreadsWait：

ThreadsWait.all_waits(threads)

这两种方式之间有什么区别吗？

我知道ThreadsWait类有其他有用的方法.
并特别询问all_waits方法.

Answer 1

documentation明确指出all_waits将在每个线程执行后执行任何传递的块;加入不提供这样的东西.

require "thwait"

threads = [Thread.new { 1 },Thread.new { 2 }]

ThreadsWait.all_waits(threads) do |t|
  puts "#{t} complete."
end # will return nil

# output:
# #<Thread:0x00000002773268> complete.
# #<Thread:0x00000002772ea8> complete.

为了实现同样的加入,我想你必须这样做：

threads.each do |t|
  t.join
  puts "#{t} complete."
end # will return threads

除此之外,all_waits方法最终调用join_nowait方法,该方法通过调用连接来处理每个线程.

没有任何阻塞,我会想象直接使用join会更快,因为你会减少导致它的所有ThreadsWait方法.所以我试了一下：

require "thwait"
require "benchmark"

loops = 100_000
Benchmark.bm do |x|
  x.report do
    loops.times do
      threads = [Thread.new { 2 * 1000 },Thread.new { 4 * 2000 }]
      threads.each(&:join)
    end
  end

  x.report do
    loops.times do
      threads = [Thread.new { 2 * 1000 },Thread.new { 4 * 2000 }]
      ThreadsWait.all_waits(threads)
    end
  end
end

# results:
# user       system     total       real
# 4.030000   5.750000   9.780000  ( 5.929623 )
# 12.810000  17.require "thwait"

threads = [Thread.new { 1 },Thread.new { 2 }]

ThreadsWait.all_waits(threads) do |t|
  puts "#{t} complete."
end # will return nil

# output:
# #<Thread:0x00000002773268> complete.
# #<Thread:0x00000002772ea8> complete.
require "thwait"

threads = [Thread.new { 1 },Thread.new { 2 }]

ThreadsWait.all_waits(threads) do |t|
  puts "#{t} complete."
end # will return nil

# output:
# #<Thread:0x00000002773268> complete.
# #<Thread:0x00000002772ea8> complete.0 29.870000 ( 17.807242 )

Ruby：等待使用join和ThreadsWait.all_waits完成的所有线程 –

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