Ruby线程搞乱变量

Question

我对红宝石很陌生，我想用线。出于目的，我希望线程能够生成线程，下面的代码如下：

require 'thread'
semaphore = Mutex.new

thr = Array.new
outputs = Array.new
scripts = Array.new
for i in 1..3
  thr[i] = Thread.new do
      puts "adding #{i} thread\n"
      puts "ready to create #{i} thread\n"
      scripts[i]= Thread.new do
        puts "in #{i} thread\n"
        puts "X#{i}\n"
        outputs[i] = "a#{i}" 
      end
  end 
end
for i in 1..3
  thr[i].join
end
for i in 1..3
  scripts[i].join
end
for i in 1..3
  puts outputs[i]
end

输出是

adding 1 thread
adding 2 thread
adding 3 thread
ready to create 3 thread
ready to create 1 thread
ready to create 1 thread
in 1 thread
in 1 thread
in 2 thread
X2
X3
X1
C:/Users/user/workspace/ruby-test/test.rb:61: undefined method `join' for nil:NilClass (NoMethodError)
    from C:/Users/liux14/workspace/ruby-test/test.rb:60:in `each'
    from C:/Users/liux14/workspace/ruby-test/test.rb:60

前三行是正确的，但下面我搞砸了。

2 i=1，1 i=2和1 i= 3，输出的I=0。

我错过了什么？

Answer 1

使用for i in 1..3语句可能会使i在for块之外可用，并使其在父线程和子线程之间共享。

试着用块来代替：

(1..3).each do |i|
  # code
end

#!/usr/bin/env ruby

require 'thread'
semaphore = Mutex.new

thr = Array.new
outputs = Array.new
scripts = Array.new
(1..3).each do |i|
  thr[i] = Thread.new do
      puts "adding #{i} thread\n"
      puts "ready to create #{i} thread\n"
      scripts[i]= Thread.new do
        puts "in #{i} thread\n"
        puts "X#{i}\n"
        outputs[i] = "a#{i}" 
      end
  end 
end
(1..3).each do |i|
  thr[i].join
end
(1..3).each do |i|
  scripts[i].join
end
(1..3).each do |i|
  puts outputs[i]
end

denis@DB:~/wk $ ./test.rb 
adding 1 thread
ready to create 1 thread
adding 3 thread
ready to create 3 thread
adding 2 thread
ready to create 2 thread
in 1 thread
X1
in 3 thread
in 2 thread
X3
X2
a1
a2
a3