[TOC] ## 4、Go是否可以无限go？ 如何限定数量？ ### 一、不控制goroutine数量引发的问题 我们都知道Goroutine具备如下两个特点 * 体积轻量 * 优质的GMP调度 那么goroutine是否可以无限开辟呢，如果做一个服务器或者一些高业务的场景，能否随意的开辟goroutine并且放养不管呢？让他们自生自灭，毕竟有强大的GC和优质的调度算法支撑？ 那么我可以先看如下一个问题。 > code1.go ```go package main import ( "fmt" "math" "runtime" ) func main() { //模拟用户需求业务的数量 task_cnt := math.MaxInt64 for i := 0; i < task_cnt; i++ { go func(i int) { //... do some busi... fmt.Println("go func ", i, " goroutine count = ", runtime.NumGoroutine()) }(i) } } ``` 结果 <img src="./pic/147-goroutines1.png" alt="image-20200328231947588" style="zoom:50%;" /> 最后被操作系统以kill信号，强制终结该进程。 ```bash signal: killed ``` 所以，我们迅速的开辟goroutine(**不控制并发的 goroutine 数量** )会在短时间内占据操作系统的资源(CPU、内存、文件描述符等)。 - CPU 使用率浮动上涨 - Memory 占用不断上涨。 - 主进程崩溃（被杀掉了） 这些资源实际上是所有用户态程序共享的资源，所以大批的goroutine最终引发的灾难不仅仅是自身，还会关联其他运行的程序。 所以在编写逻辑业务的时候，限制goroutine是我们必须要重视的问题。 --- ### 二、一些简单方法控制goroutines数量 #### 方法一：只是用有buffer的channel来限制 > code2.go ```go package main import ( "fmt" "math" "runtime" ) func busi(ch chan bool, i int) { fmt.Println("go func ", i, " goroutine count = ", runtime.NumGoroutine()) <-ch } func main() { //模拟用户需求业务的数量 task_cnt := math.MaxInt64 //task_cnt := 10 ch := make(chan bool, 3) for i := 0; i < task_cnt; i++ { ch <- true go busi(ch, i) } } ``` 结果 ```bash ... go func 352277 goroutine count = 4 go func 352278 goroutine count = 4 go func 352279 goroutine count = 4 go func 352280 goroutine count = 4 go func 352281 goroutine count = 4 go func 352282 goroutine count = 4 go func 352283 goroutine count = 4 go func 352284 goroutine count = 4 go func 352285 goroutine count = 4 go func 352286 goroutine count = 4 go func 352287 goroutine count = 4 go func 352288 goroutine count = 4 go func 352289 goroutine count = 4 go func 352290 goroutine count = 4 go func 352291 goroutine count = 4 go func 352292 goroutine count = 4 go func 352293 goroutine count = 4 go func 352294 goroutine count = 4 go func 352295 goroutine count = 4 go func 352296 goroutine count = 4 go func 352297 goroutine count = 4 go func 352298 goroutine count = 4 go func 352299 goroutine count = 4 go func 352300 goroutine count = 4 go func 352301 goroutine count = 4 go func 352302 goroutine count = 4 ... ``` 从结果看，程序并没有出现崩溃，而是按部就班的顺序执行，并且go的数量控制在了3，(4的原因是因为还有一个main goroutine)那么从数字上看，是不是在跑的goroutines有几十万个呢？  这里我们用了，buffer为3的channel, 在写的过程中，实际上是限制了速度。限制的是 ```go for i := 0; i < go_cnt; i++ { //循环速度 ch <- true go busi(ch, i) } ``` `for`循环的速度，因为这个速度决定了go的创建速度，而go的结束速度取决于 `busi()`函数的执行速度。 这样实际上，我们就能够保证了，同一时间内运行的goroutine的数量与buffer的数量一致。从而达到了限定效果。 但是这段代码有一个小问题，就是如果我们把go_cnt的数量变的小一些，会出现打出的结果不正确。 ```go package main import ( "fmt" //"math" "runtime" ) func busi(ch chan bool, i int) { fmt.Println("go func ", i, " goroutine count = ", runtime.NumGoroutine()) <-ch } func main() { //模拟用户需求业务的数量 //task_cnt := math.MaxInt64 task_cnt := 10 ch := make(chan bool, 3) for i := 0; i < task_cnt; i++ { ch <- true go busi(ch, i) } } ``` 结果 ```bash go func 2 goroutine count = 4 go func 3 goroutine count = 4 go func 4 goroutine count = 4 go func 5 goroutine count = 4 go func 6 goroutine count = 4 go func 1 goroutine count = 4 go func 8 goroutine count = 4 ``` 是因为`main`将全部的go开辟完之后，就立刻退出进程了。所以想全部go都执行，需要在main的最后进行阻塞操作。 #### 方法二：只使用sync同步机制 > code3.go ```go import ( "fmt" "math" "sync" "runtime" ) var wg = sync.WaitGroup{} func busi(i int) { fmt.Println("go func ", i, " goroutine count = ", runtime.NumGoroutine()) wg.Done() } func main() { //模拟用户需求业务的数量 task_cnt := math.MaxInt64 for i := 0; i < task_cnt; i++ { wg.Add(1) go busi(i) } wg.Wait() } ``` 很明显，单纯的使用`sync`依然达不到控制goroutine的数量，所以最终结果依然是崩溃。 结果 ```bash ... go func 7562 goroutine count = 7582 go func 24819 goroutine count = 17985 go func 7685 goroutine count = 7582 go func 24701 goroutine count = 17984 go func 7563 goroutine count = 7582 go func 24821 goroutine count = 17983 go func 24822 goroutine count = 17983 go func 7686 goroutine count = 7582 go func 24703 goroutine count = 17982 go func 7564 goroutine count = 7582 go func 24824 goroutine count = 17981 go func 7687 goroutine count = 7582 go func 24705 goroutine count = 17980 go func 24706 goroutine count = 17980 go func 24707 goroutine count = 17979 go func 7688 goroutine count = 7582 go func 24826 goroutine count = 17978 go func 7566 goroutine count = 7582 go func 24709 goroutine count = 17977 go func 7689 goroutine count = 7582 go func 24828 goroutine count = 17976 go func 24829 goroutine count = 17976 go func 7567 goroutine count = 7582 go func 24711 goroutine count = 17975 //操作系统停止响应 ``` #### 方法三：channel与sync同步组合方式 > code4.go ```go package main import ( "fmt" "math" "sync" "runtime" ) var wg = sync.WaitGroup{} func busi(ch chan bool, i int) { fmt.Println("go func ", i, " goroutine count = ", runtime.NumGoroutine()) <-ch wg.Done() } func main() { //模拟用户需求go业务的数量 task_cnt := math.MaxInt64 ch := make(chan bool, 3) for i := 0; i < task_cnt; i++ { wg.Add(1) ch <- true go busi(ch, i) } wg.Wait() } ``` 结果 ```bash //... go func 228851 goroutine count = 4 go func 228852 goroutine count = 4 go func 228853 goroutine count = 4 go func 228854 goroutine count = 4 go func 228855 goroutine count = 4 go func 228856 goroutine count = 4 go func 228857 goroutine count = 4 go func 228858 goroutine count = 4 go func 228859 goroutine count = 4 go func 228860 goroutine count = 4 go func 228861 goroutine count = 4 go func 228862 goroutine count = 4 go func 228863 goroutine count = 4 go func 228864 goroutine count = 4 go func 228865 goroutine count = 4 go func 228866 goroutine count = 4 go func 228867 goroutine count = 4 //... ``` 这样我们程序就不会再造成资源爆炸而崩溃。而且运行go的数量控制住了在buffer为3的这个范围内。 #### 方法四：利用无缓冲channel与任务发送/执行分离方式 > code5.go ```go package main import ( "fmt" "math" "sync" "runtime" ) var wg = sync.WaitGroup{} func busi(ch chan int) { for t := range ch { fmt.Println("go task = ", t, ", goroutine count = ", runtime.NumGoroutine()) wg.Done() } } func sendTask(task int, ch chan int) { wg.Add(1) ch <- task } func main() { ch := make(chan int) //无buffer channel goCnt := 3 //启动goroutine的数量 for i := 0; i < goCnt; i++ { //启动go go busi(ch) } taskCnt := math.MaxInt64 //模拟用户需求业务的数量 for t := 0; t < taskCnt; t++ { //发送任务 sendTask(t, ch) } wg.Wait() } ``` 结构 ```bash //... go task = 130069 , goroutine count = 4 go task = 130070 , goroutine count = 4 go task = 130071 , goroutine count = 4 go task = 130072 , goroutine count = 4 go task = 130073 , goroutine count = 4 go task = 130074 , goroutine count = 4 go task = 130075 , goroutine count = 4 go task = 130076 , goroutine count = 4 go task = 130077 , goroutine count = 4 go task = 130078 , goroutine count = 4 go task = 130079 , goroutine count = 4 go task = 130080 , goroutine count = 4 go task = 130081 , goroutine count = 4 go task = 130082 , goroutine count = 4 go task = 130083 , goroutine count = 4 go task = 130084 , goroutine count = 4 go task = 130085 , goroutine count = 4 go task = 130086 , goroutine count = 4 go task = 130087 , goroutine count = 4 go task = 130088 , goroutine count = 4 go task = 130089 , goroutine count = 4 go task = 130090 , goroutine count = 4 go task = 130091 , goroutine count = 4 go task = 130092 , goroutine count = 4 go task = 130093 , goroutine count = 4 ... ``` 执行流程大致如下，这里实际上是将任务的发送和执行做了业务上的分离。使得消息出去，输入SendTask的频率可设置、执行Goroutine的数量也可设置。也就是既控制输入(生产)，又控制输出(消费)。使得可控更加灵活。这也是很多Go框架的Worker工作池的最初设计思想理念。  --- 以上便是目前有关限定goroutine基础设计思路。 参考： http://team.jiunile.com/blog/2019/09/go-control-goroutine-number.html https://www.joyk.com/dig/detail/1547976674512705