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

代码语言：javascript复制

package main

import "time"

import "fmt"

func main() {
    control := make(chan interface{}, 2)
    for i := 1; i <= 10; 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: %dn", j, time.Now().Unix())
            time.Sleep(time.Second)
            <-control
        }(i)

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

代码语言：javascript复制

go func: 2, time: 1574427632
go func: 1, time: 1574427632
go func: 4, time: 1574427633
go func: 3, time: 1574427633
go func: 5, time: 1574427634
go func: 6, time: 1574427634
go func: 7, time: 1574427635
go func: 8, time: 1574427635
go func: 9, time: 1574427636
go func: 10, time: 1574427636

看时间每次只是同时执行两个

channel

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