普通网友 2018-12-17 12:10
浏览 43
已采纳

在不使用goroutines(在其作用域内)和将其传递给goroutines时,通道缓冲区如何在golang中工作?

I am absolute newbie in Golang. I am learning through Tour of Go and then implementing ideas with my own understanding. I am having problem with goroutines. I made a unbuffered channel , then sent a string to that channel.

func main() {
    p := make(chan string)
    p <- "Hello goroutine"
    fmt.Println(<-p)
}

throws error

fatal error: all goroutines are asleep - deadlock!

I get it, channel is unbuffered. (That's the reason. Right?).

But when I refactor p <- "Hello goroutine to a goroutine

func main() {
    p := make(chan string)
    go sendHello(p)
    fmt.Println(<-p)
}

func sendHello(p chan string) {
    p <- "Hello goroutine"
}

It works without problem. I read that we do not need to use pointers with maps,slices and channels with most cases to modify the value. Was channel p passed to func sendHello(p chan string) via a copy which had a separate buffer. I still can not get my head around it.

  • 写回答

1条回答 默认 最新

  • douyan1453 2018-12-17 12:27
    关注

    Keep in mind that a channel has two ends, a sender and a receiver. Your problem is about the order of execution.

    In the first example, when you use an unbuffered channel, the channel expects a receiver, while there's none as of sending Hello goroutine message, and waits until there's one (this is not the case for a buffered channel as it doesn't require to wait), and the execution never reaches the next line (i.e. a deadlock).

    But in the second example, the receiver is bound to the channel and the groutine is executed after, and both sender and receiver don't remain in a wait state.

    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 PADS Logic 原理图
  • ¥15 PADS Logic 图标
  • ¥15 电脑和power bi环境都是英文如何将日期层次结构转换成英文
  • ¥20 气象站点数据求取中~
  • ¥15 如何获取APP内弹出的网址链接
  • ¥15 wifi 图标不见了 不知道怎么办 上不了网 变成小地球了
  • ¥50 STM32单片机传感器读取错误
  • ¥15 (关键词-阻抗匹配,HFSS,RFID标签天线)
  • ¥15 机器人轨迹规划相关问题
  • ¥15 word样式右侧翻页键消失