drccfl9407 2019-03-15 03:11
浏览 17
已采纳

Goroutine存储通道值没有死锁

This code runs and ends with no deadlock error. Why?

func main() {
    ch := make(chan int)
    go func() {
        ch<-1
        ch<-2
    }()

    time.Sleep(time.Second)
}
  • 写回答

1条回答 默认 最新

  • douzengjian1535 2019-03-15 04:04
    关注

    The unbuffered channel needs two end points to work, so let's start with correct example:

    package main
    
    func main() {
        go fun2()
        <-ch
        <-ch
    }
    func fun2() {
        ch <- 1
        ch <- 2
    }
    
    var ch = make(chan int)
    

    Here fun2() sends two values and main() receives two values.


    Your sample code has only one end point so the channel is not correctly constructed, so it is deadlock, but the main goroutines exits normally so you don't see the error. Here, there is no second end point, so this is deadlock:

    package main
    
    func main() {
        var ch = make(chan int)
        ch <- 1
    }
    

    Output:

    fatal error: all goroutines are asleep - deadlock!
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

报告相同问题?

悬赏问题

  • ¥15 求差集那个函数有问题,有无佬可以解决
  • ¥15 【提问】基于Invest的水源涵养
  • ¥20 微信网友居然可以通过vx号找到我绑的手机号
  • ¥15 寻一个支付宝扫码远程授权登录的软件助手app
  • ¥15 解riccati方程组
  • ¥15 display:none;样式在嵌套结构中的已设置了display样式的元素上不起作用?
  • ¥15 使用rabbitMQ 消息队列作为url源进行多线程爬取时,总有几个url没有处理的问题。
  • ¥15 Ubuntu在安装序列比对软件STAR时出现报错如何解决
  • ¥50 树莓派安卓APK系统签名
  • ¥65 汇编语言除法溢出问题