5

sum.rb很简单。您输入两个数字，它会返回总和。

# sum.rb
puts "Enter number A"
a = gets.chomp
puts "Enter number B"
b = gets.chomp
puts "sum is #{a.to_i + b.to_i}"

robot.rb用来Open3.popen3交互sum.rb。这是代码：

# robot.rb
require 'open3'

Open3.popen3('ruby sum.rb') do |stdin, stdout, stderr, wait_thr| 
  while line = stdout.gets
    if line == "Enter number A\n"
      stdin.write("10\n")
    elsif line == "Enter number B\n"
      stdin.write("30\n")
    else
      puts line
    end
  end
end

robot.rb未能运行。似乎它卡在了sum.rb's gets.chomp。

后来我发现我必须写如下才能使它工作。您需要事先以正确的顺序输入输入。

# robot_2.rb
require 'open3'

Open3.popen3('ruby sum.rb') do |stdin, stdout, stderr, wait_thr| 
  stdin.write("10\n")
  stdin.write("30\n")
  puts stdout.read
end

让我感到困惑的是：

  1. robot_2.rb不像与 shell 交互，它更像是提供 shell 需要的东西，因为我只知道。如果一个程序需要很多输入而我们无法预测顺序怎么办？

  2. 我发现如果STDOUT.flush在每个putsin之后添加sum.rb，robot.rb可以运行。但实际上我们不能相信sum.rb作者可以添加STDOUT.flush，对吗？

谢谢你的时间！

4

1 回答 1

0

终于想出了如何做到这一点。使用write_nonblock和readpartial。您必须小心的是，stdout.readpartial它完全按照它所说的那样做，这意味着您将不得不聚合数据并gets通过查找换行符自己执行。

require 'open3'
env = {"FOO"=>"BAR", "BAZ"=>nil}
options = {}
Open3.popen3(env, "cat", **options) {|stdin, stdout, stderr, wait_thr|
    stdin.write_nonblock("hello")

    puts stdout.readpartial(4096)
    # the magic 4096 is just a size of memory from this example:
    # https://apidock.com/ruby/IO/readpartial


    stdin.close
    stdout.close
    stderr.close
    wait_thr.join
}

对于正在寻找更通用交互性（例如 ssh 交互）的人，您可能希望创建单独的线程来聚合标准输出和触发标准输入。

require 'open3'
env = {"FOO"=>"BAR", "BAZ"=>nil}
options = {}
unprocessed_output = ""
Open3.popen3(env, "cat", **options) {|stdin, stdout, stderr, wait_thr|

    on_newline = ->(new_line) do
        puts "process said: #{new_line}"
        # close after a particular line
        stdin.close
        stdout.close
        stderr.close
    end

    Thread.new do
        while not stdout.closed? # FYI this check is probably close to useless/bad
            unprocessed_output += stdout.readpartial(4096)
            if unprocessed_output =~ /(.+)\n/
                # extract the line
                new_line = $1
                # remove the line from unprocessed_output
                unprocessed_output.sub!(/(.+)\n/,"")
                # run the on_newline
                on_newline[new_line]
            end

            # in theres no newline, this process will hang forever
            # (e.g. probably want to add a timeout)
        end
    end

    stdin.write_nonblock("hello\n")

    wait_thr.join
}

顺便说一句，这不是非常线程安全的。这只是我发现的一个未优化但功能强大的解决方案，希望在未来得到改进。

于 2020-01-12T17:49:57.303 回答