Lua编程示例（八）：生产者-消费者问题

这个问题是比较经典的啦，基本所有语言的多线程都会涉及到，但是没想到Lua的这个这么复杂 抓狂
  看了好长时间才算看明白，先上个逻辑图：

   开始时调用消费者，当消费者需要值时，再调用生产者生产值，生产者生产值后停止，直到消费者再次请求。设计为消费者驱动的设计。
   图画的不太好，可以先将Filter遮住，它是过滤器对两个程序之间传递的信息进行处理。去掉Filter逻辑就更清晰些了，就是两个“线程”（其实是两个协同程序）互相调用。resume回到yield处开始，支持嵌套，返回到栈顶的yield位置。yield是非阻塞的“线程同步”。这到有点像linux里的管道通信。

function receive(prod)
 print("receive is called")
 local status,value = coroutine.resume(prod)
 return value
end

function send(x,prod)
 print("send is called")
 return coroutine.yield(x)
end

function producer()
 return coroutine.create(function ()
 print("producer is called")
 while true do
 print("producer run again")
  local x = io.read()
  send(x)
 end
 end)
end

function filter(prod)
 return coroutine.create(function ()
 for line = 1,1000 do
  print("enter fliter "..line)
  local x = receive(prod)
  print("receive in filter finished")
  x= string.format("%5d %s",line,x)
  send(x,prod)
 end
 end)
end

function consumer(prod)
 print("consumer is called")
 while true do
 print("consumer run again")
 local x = receive(prod)
 print("retrun customer")
 io.write(x,"\n")
 end
end

p = producer()
f=filter(p)
consumer(f)

运行结果：

consumer is called
consumer run again
receive is called
enter fliter 1
receive is called
producer is called
producer run again
fsy
send is called
receive in filter finished
send is called
retrun customer
  1 fsy
consumer run again
receive is called
enter fliter 2
receive is called
producer run again
gaga
send is called
receive in filter finished
send is called
retrun customer
  2 gaga
consumer run again
receive is called
enter fliter 3
receive is called
producer run again
......