douyi3760 2017-12-15 23:38
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I'm new with golang channels and could not underestimate why does program behavior changes if I specify channel buffer size

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")
}

Output is:

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

If I replace in := make(chan bool) with in := make(chan bool, 1) It works properly. Why does it happen?

  • 写回答

1条回答 默认 最新

  • dongma2388 2017-12-15 23:45
    关注

    This is because the main goroutine is stuck writing to in

    in <- true
    

    and the other goroutine is stuck writing to out

    out <- 1
    

    If you make the write to out a case aswell it should work:

    for {
        select {
        case <-in:
            fmt.Println("Close")
            close(out)
            return
        case out <- 1:
            fmt.Println("Out")
            break
        }
    }
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

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