dsp15140275697 2014-08-28 02:14
浏览 10
已采纳

进入频道,似乎还可以,但是却陷入僵局

package main
import "fmt"
import "time"

func main() {
     c := make(chan int)
     c <- 42    // write to a channel

     val := <-c // read from a channel
     println(val)
}

I think c <- 42 put 42 to channel c, then val := <-c put value in c to val. but why does it get deadlock?

  • 写回答

1条回答 默认 最新

  • duanba8173 2014-08-28 02:22
    关注

    You have created an unbuffered channel. So the statement c <- 42 will block until some other goroutine tries to receive a value from the channel. Since no other goroutine is around to do this, you got a deadlock. There are two ways you could fix this:

    1. Perform the receive in a different goroutine.
    2. Add a buffer to the channel. For example, c := make(chan int, 1) would allow you to send a single value on the channel without blocking.
    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

  • ¥50 求解vmware的网络模式问题
  • ¥24 EFS加密后,在同一台电脑解密出错,证书界面找不到对应指纹的证书,未备份证书,求在原电脑解密的方法,可行即采纳
  • ¥15 springboot 3.0 实现Security 6.x版本集成
  • ¥15 PHP-8.1 镜像无法用dockerfile里的CMD命令启动 只能进入容器启动,如何解决?(操作系统-ubuntu)
  • ¥30 请帮我解决一下下面六个代码
  • ¥15 关于资源监视工具的e-care有知道的嘛
  • ¥35 MIMO天线稀疏阵列排布问题
  • ¥60 用visual studio编写程序,利用间接平差求解水准网
  • ¥15 Llama如何调用shell或者Python
  • ¥20 谁能帮我挨个解读这个php语言编的代码什么意思?