Question

我是golang频道的新手，如果指定通道缓冲区大小，我不能低估为什么程序行为会发生变化

package main

import (
    "fmt"
)

func channels(in <-chan bool, out chan int) {
    for {
        select {
        case _ = <-in:
            fmt.Println("Close")
            close(out)
            return
        default:
            fmt.Println("Out")
            out <- 1
            break
        }
    }
}

func main() {
    in := make(chan bool)
    // in := make(chan bool, 1)
    out := make(chan int)

    go channels(in, out)

    i := 0

    // Loop:
    for n := range out {
        fmt.Println(n)

        i += 1

        if i > 10 {
            fmt.Println("Send close")
            in <- true // Here it is became asleep
            return
        }
    }

    fmt.Println("Done")
}

输出是：

Out
Out
1
1
Out
Out
1
1
Out
Out
1
1
Out
Out
1
1
Out
Out
1
1
Out
Out
1
Send close
fatal error: all goroutines are asleep - deadlock!

goroutine 1 [chan send]:
main.main()
    /home/user/go-lessons/3/chan_4.go:39 +0x2c3

goroutine 19 [chan send]:
main.channels(0xc820072060, 0xc8200720c0)
    /home/user/go-lessons/3/chan_4.go:16 +0x241
created by main.main
    /home/user/go-lessons/3/chan_4.go:27 +0x97
exit status 2

如果我将in := make(chan bool)替换为in := make(chan bool, 1)它可以正常运行。为什么会这样？

Answer 1

这是因为主要的goroutine被写入in

in <- true

而另一个goroutine被写入out

out <- 1

如果您要写出case，那么它应该有效：

for {
    select {
    case <-in:
        fmt.Println("Close")
        close(out)
        return
    case out <- 1:
        fmt.Println("Out")
        break
    }
}

Golang频道睡着了

1 个答案: