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 ......
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