[Go] 利用有缓存channel控制同时并发的数量

如果有一个大循环,里面每一个都开启groutine,那么瞬间就会开启非常多的groutine,要解决这个问题就要用channel的阻塞特性来解决

package main

import "time"

import "fmt"

func main() {
    control := make(chan interface{}, )
    for i := ; i <= ; i++ {
        control <- i //这里应该放上面,如果放下面就会每次都执行三个了
        go func(j int) {
            fmt.Printf("go func: %d, time: %d\n", j, time.Now().Unix())
            time.Sleep(time.Second)
            <-control
        }(i)

    }
    //主groutine不要断
    for {
        time.Sleep(time.Second)
    }
}

go func: , time:
go func: , time:
go func: , time:
go func: , time:
go func: , time:
go func: , time:
go func: , time:
go func: , time:
go func: , time:
go func: , time: 

看时间每次只是同时执行两个